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2015-01-20T00:03:03Z alexander-01: can anyone point me to documentation on opening and writing to a binary file .. thanks
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2015-01-20T00:26:31Z pjb: alexander-01: there's nothing in the r5rs standard. http://www.schemers.org/Documents/Standards/R5RS/HTML/r5rs-Z-H-9.html#%_sec_6.6  You will have to look at your implementation documentation, or perhaps in the r7rs standard.
2015-01-20T00:26:31Z rudybot: http://tinyurl.com/3f2se39
2015-01-20T00:26:39Z pjb: lrs: I will. Good night.
2015-01-20T00:28:46Z alexander-01: pjb: thanks
2015-01-20T00:32:15Z kongtomorrow: pjb: I did the other person's problem just 'cause I kind of felt like it. Any particular comments? https://gist.github.com/kongtomorrow/5cf71dd76151293eb976
2015-01-20T00:35:43Z pjb: kongtomorrow: it's ok.  I would still advise to replace car/cadr by giving them names, eg. (define rule-antecendent car) (define rule-result-type cadr).
2015-01-20T00:38:03Z kongtomorrow: pjb: 👍
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2015-01-20T08:31:28Z zacts: hi
2015-01-20T08:31:31Z zacts: I <3 scheme
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2015-01-20T11:09:17Z hellofunk: zacts: yeah, you do!
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2015-01-20T15:38:55Z learningmore: what should one go over first
2015-01-20T15:38:59Z learningmore: SICP or HTDP?
2015-01-20T15:41:10Z z0d: they serve different purposes, but HTDP is easier
2015-01-20T15:42:21Z pjb: To learn how to program, sicp. To learn scheme, htdp.
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2015-01-20T15:42:50Z z0d: SICP is more about computer science IMO
2015-01-20T15:43:43Z learningmore: trying to use MIT-GNU Scheme along with SICP, but the Emacs commands are bugging me a bit
2015-01-20T15:43:56Z learningmore: is there a better REPL I could be using?
2015-01-20T15:45:15Z pjb: Did you use M-x run-scheme RET ?
2015-01-20T15:46:49Z learningmore: no
2015-01-20T15:48:56Z wasamasa: you should
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2015-01-20T19:26:13Z lrs_: pjb, There?
2015-01-20T19:27:32Z pjb: yes
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2015-01-20T19:27:55Z lrs_: pjb, Do you feel like lecturing me with the problem i had before?
2015-01-20T19:28:08Z pjb: ok.
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2015-01-20T19:30:38Z lrs_: pjb, I was thinking of doing the quotient, /, * operators
2015-01-20T19:30:45Z lrs_: But Im not really sure how t ostart
2015-01-20T19:31:04Z pjb: Just add rules to the rules list.
2015-01-20T19:31:45Z lrs_: pjb, What was the rules list?
2015-01-20T19:31:51Z pjb: (define rules '( (--> (quotient int non-zero-int) int)   (--> (/ int non-zero-int) real) …)
2015-01-20T19:31:53Z pjb: )
2015-01-20T19:32:10Z lrs_: pjb, I dont get what that means
2015-01-20T19:32:14Z lrs_: What you just wrote
2015-01-20T19:32:35Z lrs_: Mainly the '( and -->
2015-01-20T19:32:35Z pjb: it means that when you (check '(quotient int non-zero-int)) you get int as result.
2015-01-20T19:32:47Z pjb: 'x = (quote x)
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2015-01-20T19:33:18Z pjb: (quote x) --> x
2015-01-20T19:33:20Z lrs_: pjb http://lpaste.net/118869
2015-01-20T19:33:27Z pjb: quote returns its argument without evaluating it.
2015-01-20T19:33:28Z lrs_: Thats how it look like right now
2015-01-20T19:33:38Z pjb: it is used to introduce literals.
2015-01-20T19:34:13Z lrs_: pjb, Whats literals
2015-01-20T19:34:51Z pjb: At that point, I would advise to pause with sicp, and use another book to learn some scheme programming first.
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2015-01-20T19:35:41Z pjb: lrs_: you should consider studing http://htdp.org/2003-09-26/ first.
2015-01-20T19:35:53Z lrs_: pjb, Well, Im swedish so I have ahard time with difficult english words
2015-01-20T19:36:08Z pjb: ok.
2015-01-20T19:36:21Z pjb: literals are data that are present literally in the source code of the program.
2015-01-20T19:36:56Z pjb: Now, in lisp, it is a little difficult to differentiate data from code, since we write lisp code  using the syntax defined to write lisp data: sexp (symbolic expressions).
2015-01-20T19:36:56Z lrs_: So, it is stuff that is preset in scheme?
2015-01-20T19:37:23Z pjb: So to prevent the evaluator to evaluate the data that is really data and that should not be considered code, we use the quote operator.
2015-01-20T19:37:35Z pjb: Yes.
2015-01-20T19:37:49Z pjb: lrs_: you should also consider reading r5rs.  This is less than 50 pages.
2015-01-20T19:37:50Z lrs_: Ah, ok. I figured it was something like that. But when you say introduce literals?
2015-01-20T19:37:55Z lrs_: pjb, Yeah, I have
2015-01-20T19:38:01Z lrs_: I have it printed out
2015-01-20T19:38:17Z pjb: Well, when I say reading, I mean memorizing. Not priting it out.
2015-01-20T19:38:23Z lrs_: Oh, ok
2015-01-20T19:39:00Z lrs_: Well, when I said that I meant. "I know what it is and I have read *some* of it"
2015-01-20T19:39:39Z pjb: Since r5rs is less than 50 pages, it should be read and know entirely.
2015-01-20T19:41:04Z lrs_: pjb, Well, I know most of it
2015-01-20T19:41:21Z lrs_: But are you suggesting that there is some chapter there that is beneficial to me
2015-01-20T19:42:49Z pjb: 100% of r5rs.  Perhaps not the whole htdp, because it's big, and you may want to come back to sicp before finishing htdp, but you should read enough of it to know some basics such as what a literal is and what quote is used for.
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2015-01-20T19:45:02Z lrs_: pjb, Oh. Ok, I get know what you mean. Of course I know what ' i.e quote means
2015-01-20T19:45:27Z pjb: What occured in 1959, is that Steve Russell, a student of John McCarthy, implemented the eval function from the AIM-8 paper, and therefore, used S-expr to write programs which were passed as data to the eval function.
2015-01-20T19:45:46Z pjb: And it just worked, S-expr data were programs when interpreted by eval.
2015-01-20T19:46:35Z pjb: This is important, because it shows that the notion of program, depends on the processor you have for this class of programs.
2015-01-20T19:47:02Z pjb: The rules  we are defining are actually a program, for the processor named apply-rules.
2015-01-20T19:47:28Z lrs_: Right
2015-01-20T19:49:23Z lrs_: pjb, But what does quote and literals has to to with our program rule
2015-01-20T19:50:01Z pjb: Since you are writing it in a file named something.scm, it will be interpreted by a scheme processor!
2015-01-20T19:50:16Z pjb: Therefore if you don't quote it, it will be taken as lisp (scheme) program.
2015-01-20T19:50:37Z lrs_: You mean if I use define?
2015-01-20T19:50:43Z lrs_: I need to use quote instead of define?
2015-01-20T19:50:44Z pjb: But --> is not a scheme operator, and ((--> (a i) i) (--> (b r) r)) is not a valid scheme form.
2015-01-20T19:51:13Z pjb: However you use it, you cannot write ((--> (a i) i) (--> (b r) r)) in a scheme program while --> is not an adequately defined scheme operator.
2015-01-20T19:51:31Z lrs_: Yes, I get that
2015-01-20T19:52:12Z pjb: Therefore we quote it, and then it becomes lisp data. (that will be used as apply-rules program).
2015-01-20T19:53:06Z lrs_: Ok
2015-01-20T19:53:35Z pjb: And keeping the list of rules in a list bound to a variable named rules is logical and practical.
2015-01-20T19:54:03Z pjb: It's better to name the variable rules than schmiblick, because we can then easily infer that this variable holds rules.
2015-01-20T19:54:05Z lrs_: So I start with writing in my something.scm , (quote (quotient int non-zero-int) int)
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2015-01-20T19:54:33Z pjb: And having the rules in a list bound to a variable allows us to update this list easily, and even not necessarily in one place.
2015-01-20T19:54:54Z pjb: lrs_: no, you start by reading htdp.
2015-01-20T19:55:38Z lrs_: Where would I read the thing that would benefit me?
2015-01-20T19:55:46Z lrs_: I have studied scheme for 4-5 months
2015-01-20T19:56:11Z lrs_: I dont really feel like spending alot of hours on a book if it is jus old news
2015-01-20T19:56:22Z lrs_: I need to find stuff that helps me with this particular problem..
2015-01-20T19:56:24Z pjb: You are asking very naive questions, so I doubt you've studied scheme programming for 4 months.
2015-01-20T19:57:23Z lrs_: Naive how?
2015-01-20T19:57:37Z pjb: lrs_: let's start by the apply-rules function.  If you've studied scheme programming for 4 months, could you tell me what apply-rules does?
2015-01-20T19:58:38Z lrs_: In what sense?
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2015-01-20T19:58:57Z lrs_: What it does in the chain of programs?
2015-01-20T19:59:00Z lrs_: OR what its purpose is
2015-01-20T19:59:24Z pjb: Could you describe in english the purpose of apply-rules?
2015-01-20T19:59:27Z lrs_: As in pre and post
2015-01-20T19:59:31Z pjb: Yes.
2015-01-20T20:00:04Z lrs_: It takes rules and applies them
2015-01-20T20:00:31Z pjb: More or less.
2015-01-20T20:01:16Z pjb: It searches for a rule, and if it finds it, then it applies only this rule and returns the result of this rule application.
2015-01-20T20:01:24Z lrs_: My main problem is basically that 1. I dont know how I should define int, bool, *, quotient. 2. I dont know how to structure up all the chains of programs 3. I dont know how I should iterate through something like this > (check '(= (< (+ int int) (quotient int int)) (> int int)))
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2015-01-20T20:01:58Z pjb: 2. this "iteration" is actually a recursion, and it's performed by the check procedure.
2015-01-20T20:02:01Z lrs_: Thats my problem
2015-01-20T20:02:07Z lrs_: Sorry, recursion
2015-01-20T20:02:10Z pjb: It's the (map check args) call.
2015-01-20T20:02:32Z lrs_: YEs
2015-01-20T20:02:36Z lrs_: That I understand
2015-01-20T20:02:47Z pjb: We will be able to see that by adding a few display in the procedures when we'll have a complete program.
2015-01-20T20:03:08Z lrs_: Yes
2015-01-20T20:03:13Z pjb: 1. the definition of int bool * quotient, etc is given implicitely by the rules where they are present.
2015-01-20T20:03:21Z pjb: hence the importance of writing the rules list.
2015-01-20T20:04:36Z lrs_: Yes
2015-01-20T20:04:41Z lrs_: And how do I do that
2015-01-20T20:04:58Z lrs_: You said i need to use quote, because otherwise scheme interprets it as a program
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2015-01-20T20:05:04Z pjb: But before, can you explain HOW apply-rules does what it does?
2015-01-20T20:06:15Z lrs_: How do you mean how? In what sense?
2015-01-20T20:06:29Z pjb: explaining the code of apply-rules.
2015-01-20T20:06:32Z pjb: What does it does?
2015-01-20T20:06:38Z pjb: How does it do it?
2015-01-20T20:06:42Z lrs_: (define (apply-rules operator arg-types) (find-if operator arg-types) real)
2015-01-20T20:06:46Z lrs_: Is how it looks like right now
2015-01-20T20:06:56Z pjb: Right. Can you explain it in English?
2015-01-20T20:07:37Z pjb: This definition is wrong.  There's another definition below.  You have duplicate definition in your scheme source file.  This is not allowed by the scheme standard.
2015-01-20T20:07:57Z lrs_: It takes a operator and applies it to arg-types? And then uses find-if operator is a arg-types.. and the ngive real. I really need yo uto explain what you want this to do because I dont understand it. And also (find-if predicate? list) so I dont really know what it does
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2015-01-20T20:09:19Z lrs: Sorry
2015-01-20T20:09:22Z lrs: This was some old stuff
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2015-01-20T20:09:39Z lrs: pjb, http://lpaste.net/118875 Is this the apply rule you are talking about?
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2015-01-20T20:10:43Z pjb: lrs: Have a look at this: http://paste.lisp.org/+3440
2015-01-20T20:11:06Z pjb: Better use paste.lisp.org for pasting lisp/scheme code.
2015-01-20T20:11:23Z lrs: Ok
2015-01-20T20:11:34Z pjb: lrs:  notice the newlines I inserted.  DrRacket didn't insert them as you said yesterday…
2015-01-20T20:11:48Z pjb: lrs: now can you see better what apply-rules does?
2015-01-20T20:11:58Z lrs: Yeah, sorry. I meant that when you press shift it just oders them correctly
2015-01-20T20:12:06Z pjb: What's the operator of the expression in the body of apply-rules?
2015-01-20T20:12:08Z lrs: IT doesnt make newlines correct so youre right
2015-01-20T20:12:51Z lrs: pjb, YEs. Maybe I have a bit difficult understanding what matchingrule does and the lambda inside the function does
2015-01-20T20:13:06Z pjb: lrs: you are looking too deep.
2015-01-20T20:13:14Z pjb: reading lisp is very easy, but you should not skip steps.
2015-01-20T20:13:25Z pjb: first you have to look at the first element of the toplevel sexp.
2015-01-20T20:13:41Z lrs: Check?
2015-01-20T20:14:06Z pjb: So, it's define.  The first argument is (apply-rules operator arg-types) which means that we are defining a function named apply-rules, with two arguments, named operator and arg-types.
2015-01-20T20:14:13Z pjb: Then follows the body of the function.
2015-01-20T20:14:32Z pjb: the body of apply-rules has only one expression. What's the operator of the expression in that body?
2015-01-20T20:15:37Z lrs: Sorry, my english is bad. What do you mean when you say expression
2015-01-20T20:15:55Z lrs: Do you mean let?
2015-01-20T20:15:58Z pjb: expression = s-expression = atom or list of s-expression.
2015-01-20T20:16:00Z pjb: Yes, it's let.
2015-01-20T20:16:03Z pjb: What does let do?
2015-01-20T20:16:30Z lrs: Evaluates
2015-01-20T20:16:44Z pjb: what else?
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2015-01-20T20:17:39Z pjb: lrs: you have the right to cheat, and to go read the r5rs to answer my questions!
2015-01-20T20:17:40Z pjb: http://www.schemers.org/Documents/Standards/R5RS/HTML/r5rs-Z-H-7.html#%_idx_124
2015-01-20T20:17:40Z rudybot: http://tinyurl.com/695ay7
2015-01-20T20:18:21Z lrs: ömb
2015-01-20T20:18:24Z lrs: Woops
2015-01-20T20:18:30Z lrs: pjb, You mean it gives values to things?
2015-01-20T20:18:35Z pjb: Yes.
2015-01-20T20:18:52Z pjb: That's its purpose in life, to bind values to variables.
2015-01-20T20:19:01Z lrs: Correct
2015-01-20T20:19:44Z pjb: In the semantic section of let, in r5rs, you see that it does several things:  evaluates the init forms first.  Bind the results of those inits to their corresponding variables,  then it evaluates the body expressions and finally it returns the result of the last expression of the body.
2015-01-20T20:19:54Z pjb: So our let in apply-rules, what does it do?
2015-01-20T20:20:06Z pjb: How many variables does it bind?
2015-01-20T20:21:07Z lrs: pjb, one? rule?
2015-01-20T20:21:16Z pjb: one, named matching-rule
2015-01-20T20:21:35Z pjb: Notice that the semantics gives order of processing: eval init(s), bind variable(s), eval body, return result.
2015-01-20T20:21:35Z lrs: Ah, right
2015-01-20T20:21:42Z pjb: But I ask first what variable(s) is bound.
2015-01-20T20:21:59Z pjb: That's because variables serve as intermediary between the init(s) forms and the body forms.
2015-01-20T20:22:30Z pjb: And with their names, they can give good hint of the meaning of the the values they're bound to. (and therefore of the meaning of the init(s) forms).
2015-01-20T20:22:56Z pjb: So you can often read a let form without reading the details of the init(s) form, just by reading the name of the variables, and the body forms.
2015-01-20T20:23:21Z pjb: But of course, you may enter into the details and see what the init(s) forms are , which we're going to do now.
2015-01-20T20:23:37Z pjb: How is matching-rule initialized?  What's the operator of its init form?
2015-01-20T20:24:40Z lrs: matching rule is initzialized by let?
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2015-01-20T20:25:21Z pjb: Well, let binds the result of the init form to the variable matching-rule.  But this init form, which is evaluated by let, what is it? What's its operator?
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2015-01-20T20:25:46Z lrs: The operator of let?
2015-01-20T20:25:54Z pjb: It's find-if
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2015-01-20T20:26:16Z pjb: The init form for matching-rule, is the expressions just after matching-rule in the binding list.
2015-01-20T20:26:18Z lrs: I have a bit trouble understanding what you mean with init form
2015-01-20T20:26:32Z pjb: Check the syntax of let in the r5rs page linked above.
2015-01-20T20:26:53Z lrs: Ah, ok
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2015-01-20T20:27:01Z pjb: (let <bindings> <body>)    bindings ::= ((<variable1> <init1) …)
2015-01-20T20:27:06Z pjb: (let <bindings> <body>)    bindings ::= ((<variable1> <init1>) …)
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2015-01-20T20:27:26Z lrs: Yes, well then it naturally is matching-rule. I get now what you mean
2015-01-20T20:28:15Z pjb: the variable is matching-rule. The init form of the matching-rule variable is (find-if …).  The operator of the init form of the matching-rule variable is find-if.
2015-01-20T20:28:21Z pjb: What is find-if?
2015-01-20T20:28:46Z pjb: procedure, macro, syntax ?
2015-01-20T20:29:08Z bjz quit (Ping timeout: 245 seconds)
2015-01-20T20:29:27Z lrs: Find-if is a function that checks if a list is a predicate or not
2015-01-20T20:29:45Z lrs: find-if predicate? lst
2015-01-20T20:29:54Z pjb: not exactly.  it's a procedure (in scheme terminology), or a function (in general lisp terminology).
2015-01-20T20:30:05Z lrs: Right, a procedure
2015-01-20T20:30:11Z pjb: It finds the first element in the list for which the predicate? function returns true.
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2015-01-20T20:30:48Z pjb: I'm more of a lisper than a schemer, so I often use function instead of procedure. Furthermore, since I learned pascal, procedures don't return values, while functions do ;-)
2015-01-20T20:31:10Z pjb: Ok, this find-if function is not a standard function. You will have to write it.
2015-01-20T20:31:17Z pjb: How many arguments does it take?
2015-01-20T20:31:17Z lrs: Thats right
2015-01-20T20:31:26Z lrs: 2
2015-01-20T20:31:31Z pjb: What's the first argument?
2015-01-20T20:31:36Z lrs: predicate?
2015-01-20T20:31:42Z pjb: Yes.  What is a predicate?
2015-01-20T20:31:55Z lrs: predicate? checks if something is a predicate or not
2015-01-20T20:32:39Z lrs:  Its a boolean basically
2015-01-20T20:32:53Z lrs: If something is this then it is that
2015-01-20T20:33:17Z pjb: Yes but no :-)  A predicate is a function that returns a boolean determined by an object.  Predicate functions are often named with a question mark at the end of their names.  But here, we write predicate? as parameter name with the question mark, which is a little redundant.
2015-01-20T20:33:34Z pjb: predicate? will be the name of the parameter, but it will be bound to a function passed as argument to find-if.
2015-01-20T20:33:51Z pjb: A predicate is a function returning a boolean, not a mere boolean.
2015-01-20T20:33:54Z pjb: Ok.
2015-01-20T20:34:00Z pjb: What's the second argument to find-if?
2015-01-20T20:35:33Z lrs: list
2015-01-20T20:35:34Z lrs: lst
2015-01-20T20:35:48Z pjb: Yes, it will be a list of objects.
2015-01-20T20:36:05Z pjb: Now, in the case of matching-rule, what is this second argument to find-if?
2015-01-20T20:36:27Z lrs: The whole lambda functon
2015-01-20T20:36:44Z pjb: this would be the first argument.  The lambda function is our predicate.
2015-01-20T20:36:49Z pjb: What about the second argument?
2015-01-20T20:37:02Z lrs:  Oh, sorry, rules
2015-01-20T20:37:21Z pjb: Right.  So rules will have to be a variable bound to a list of rules.
2015-01-20T20:37:40Z pjb: Our lambda function has as parameter a rule, a single element of the list bound to rules.
2015-01-20T20:37:51Z pjb: Ok?
2015-01-20T20:38:27Z lrs: Ah ok
2015-01-20T20:38:54Z pjb: Let's look this lambda predicate.  What's the operator of the expression in its body?
2015-01-20T20:39:09Z lrs: and
2015-01-20T20:39:24Z pjb: Ok, so we can assume it will return a boolean, which is good, since it is a predicate :-)
2015-01-20T20:39:34Z lrs: Yes
2015-01-20T20:39:37Z pjb: What does this test mean (the whole and form)?
2015-01-20T20:40:27Z lrs: IT checks if the first and second argument in apply-rules is eq to the rule-iperator and that arg-types is equalt to rule-argument-types rule
2015-01-20T20:40:39Z pjb: Good.
2015-01-20T20:41:02Z pjb: Notice that in the paste, there's a bug, since eq? is used to both.  We want to use equal? for the arg-types, since it's a list.
2015-01-20T20:41:20Z pjb: eq? would test whether the two lists are actually the same identical list.
2015-01-20T20:41:23Z lrs: YEs, so I figured. eq? is for numerical things i think?
2015-01-20T20:41:27Z lrs: Oh
2015-01-20T20:41:31Z pjb: No. For numbers use =
2015-01-20T20:41:35Z lrs: Right
2015-01-20T20:41:46Z pjb: eq? should be avoided actually. It's better in general to use eql?
2015-01-20T20:41:55Z pjb: equal? will test whether the two lists contain the same elements.
2015-01-20T20:42:21Z pjb: Ok. so that means the find-if will find a rule that has the operator and arg-types.
2015-01-20T20:42:33Z pjb: Let's look at the body of the let form now.
2015-01-20T20:42:37Z pjb: What's the operator?
2015-01-20T20:43:06Z lrs: matching-rule?
2015-01-20T20:43:23Z pjb: This is the variable bound by let.  What's in the body of the let form?
2015-01-20T20:44:29Z lrs: Everything except the if statement?
2015-01-20T20:44:45Z pjb: On the contrary. The body of the let form contains only this if expression.
2015-01-20T20:44:53Z pjb: In lisp, there are no statement, only expressions.
2015-01-20T20:45:33Z lrs: Right
2015-01-20T20:45:48Z pjb: Once let has evaluated the init form, the find-if call, it will bind the result of find-if (hopefully a rule) to the variable matching-rule, and then evaluate its body, this if form.
2015-01-20T20:45:54Z pjb: What do we test with this if?
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2015-01-20T20:48:04Z lrs: pjb, Right. If matching-rule is nil. Then it gives an error. Otherwise.. Im not sure.. It does something with rule-result-type
2015-01-20T20:48:32Z pjb: ok.  Remember that let returns as result, the result of the last expression in its body.
2015-01-20T20:48:46Z lrs: Aha
2015-01-20T20:48:51Z pjb: Since we have only one expression, this if form, then it's the result of this if form that will become the result of the let form.
2015-01-20T20:48:53Z lrs: So the last line
2015-01-20T20:49:01Z lrs: Is a way to save the stuff
2015-01-20T20:49:01Z lrs: ?
2015-01-20T20:49:18Z pjb: In the case of error, it won't return.  In the other case, it will return the (rule-result-type matching-rule).
2015-01-20T20:49:28Z pjb: It's not a "saving" it's a _returning_.
2015-01-20T20:49:52Z pjb: (let ((a 0))  (+ (if (= 0 a) 1 2) 3)) --> 4
2015-01-20T20:49:55Z lrs: It returns the value of matching-rule to rule-result?
2015-01-20T20:50:14Z pjb: Since there are only expressions, an expression like if has a value, a result.
2015-01-20T20:50:42Z pjb: No rule-result-type is (will be) a function taking matching-rule as argument.
2015-01-20T20:50:52Z pjb: And this function will return a result, the result type of the rule.
2015-01-20T20:51:12Z pjb: remember the rule format: (--> (operator . argument-types) result-type)
2015-01-20T20:52:01Z pjb: Now, we have all those functions that are undefined: rule-operator rule-argument-types rule-result-type find-if ;  you should start by implementing the rule-… functions.  Annotate the lisp paste.
2015-01-20T20:53:14Z lrs: Annotate = Comment ?
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2015-01-20T20:54:00Z lrs: Ok, Im gonna try to write some crappy code
2015-01-20T20:54:05Z pjb: It's more like a new version of the paste.  Click on the Annotate button, and you can add your code.
2015-01-20T20:55:06Z lrs: pjb, Do I need to alter "rule" in some way
2015-01-20T20:55:25Z pjb: No.
2015-01-20T20:55:46Z pjb: In general, we avoid, or just don't need to modify variables or data structures.
2015-01-20T20:56:01Z lrs: Ok, so ill only focus on the ones you mentioned. Got it
2015-01-20T20:56:03Z pjb: By avoiding mutation, we can avoid a lot of bugs.
2015-01-20T21:03:21Z lrs: pjb, This is the hardest part
2015-01-20T21:03:34Z pjb: Ok, let's consider rule-operator.
2015-01-20T21:03:44Z pjb: Can you give me a rule as an example?
2015-01-20T21:04:10Z lrs: > (+ int int)
2015-01-20T21:04:10Z lrs: int
2015-01-20T21:04:28Z pjb: The format of a rule would be: (--> (operator . argument-types) result-type)
2015-01-20T21:04:38Z pjb: So an example of a rule would be: (--> (+ int int) int)
2015-01-20T21:04:40Z pjb: right?
2015-01-20T21:04:48Z lrs: Yes
2015-01-20T21:05:00Z pjb: So what should (rule-operator '(--> (+ int int) int)) return?
2015-01-20T21:05:38Z lrs: Error? (+ int int) int) doesnt work
2015-01-20T21:05:49Z lrs: Or wait
2015-01-20T21:05:59Z lrs: Sorry
2015-01-20T21:06:00Z lrs: #t
2015-01-20T21:06:01Z pjb: the operator in the rule (--> (+ int int) int) is +
2015-01-20T21:06:09Z lrs: Oh
2015-01-20T21:06:12Z pjb: the operator in the rule (--> (* int int) int) is *
2015-01-20T21:06:13Z lrs: Ah, yes
2015-01-20T21:06:14Z pjb: etc.
2015-01-20T21:06:22Z lrs: But why  int at the end?
2015-01-20T21:06:22Z pjb: How can we get this operator from the rule?
2015-01-20T21:06:31Z pjb: the int at the end is the result-type.
2015-01-20T21:06:37Z pjb: rule format: (--> (operator . argument-types) result-type)
2015-01-20T21:06:53Z lrs: Why is result-type conencted to rule operator
2015-01-20T21:07:22Z pjb: because the meaning of the rule, is that when we apply the operator to arguments of the given type, we obtain a value of the result type.
2015-01-20T21:07:30Z pjb: given argument types
2015-01-20T21:08:05Z pjb: The purpose of the rules, is to memorize the correspondance between the operator argument-types tuples, and the result types.
2015-01-20T21:08:16Z pjb: --> is only there to make it beautiful
2015-01-20T21:08:29Z pjb: Notice that we could choose another format, such as: ((+ int int) --> int)
2015-01-20T21:09:02Z pjb: It's just because we are used to Polish notation in lisp that I put the --> first.
2015-01-20T21:09:34Z pjb: But there would be good arguments to choose this ((+ int int) --> int) format.
2015-01-20T21:10:02Z pjb: That's the point of using rule-operator rule-argument-types and rule-result-type functions: this will let us change the format easily.
2015-01-20T21:10:23Z pjb: So, given (--> (* int int) int), how can we obtain the *?
2015-01-20T21:10:37Z pjb: So, given a list such as: (--> (* int int) int), how can we obtain the *?
2015-01-20T21:12:06Z lrs: if operator equal * -> save it in some way
2015-01-20T21:12:28Z pjb: No, don't think "saving". Saving is mutating some variable.
2015-01-20T21:12:30Z pjb: We don't need that.
2015-01-20T21:12:41Z lrs: Yeah I knw
2015-01-20T21:12:44Z pjb: We have functions to cut lists: car and cdr.
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2015-01-20T21:12:58Z pjb: What is (car '(--> (* int int) int)) ?
2015-01-20T21:13:04Z lrs: Ah, right
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2015-01-20T21:13:39Z lrs: *
2015-01-20T21:13:52Z pjb: (car '(--> (* int int) int)) returns --> ; says my scheme.
2015-01-20T21:13:59Z pjb: So it's not good.
2015-01-20T21:14:11Z pjb: What about (cdr '(--> (* int int) int)) ?
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2015-01-20T21:14:25Z lrs: (1 1)
2015-01-20T21:14:30Z lrs: I dont really understand the --> thing
2015-01-20T21:14:36Z pjb: --> is a symbol.
2015-01-20T21:14:47Z lrs: A procedure symbol?
2015-01-20T21:14:52Z pjb: like gives or rule or thingy.
2015-01-20T21:14:55Z pjb: No, just a symbol.
2015-01-20T21:14:57Z lrs: Ok
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2015-01-20T21:15:10Z pjb: Lisp is a symbolic language.
2015-01-20T21:15:20Z pjb: (cdr '(--> (* int int) int)) returns: ((* int int) int)
2015-01-20T21:15:24Z pjb: not good, but it's better.
2015-01-20T21:15:34Z pjb: What about (car (cdr '(--> (* int int) int))) ?
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2015-01-20T21:16:36Z lrs: Oh
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2015-01-20T21:16:56Z lrs: But (cdr '(* int int)) gives (int int)
2015-01-20T21:17:24Z pjb: Yes, but let's build a complete expression, starting from the rule.
2015-01-20T21:17:35Z pjb: You should evaluate those things in the REPL of course.
2015-01-20T21:17:49Z pjb: We could bind the rule to a variable:
2015-01-20T21:18:20Z pjb: (let  ((rule '(--> (* int int) int)))   (car (cdr rule))) returns: (* int int)
2015-01-20T21:18:29Z pjb: better, but not yet.
2015-01-20T21:18:41Z lrs: yes
2015-01-20T21:18:54Z pjb: So what could we use instead of ??? to get the *? (let  ((rule '(--> (* int int) int)))   (??? (car (cdr rule))))
2015-01-20T21:19:38Z lrs: (cdr (car))
2015-01-20T21:20:04Z pjb: If you use copy and paste, you can easily answer complete expressions.
2015-01-20T21:20:14Z pjb: (let  ((rule '(--> (* int int) int)))   (car (car (cdr rule)))) returns: * ; yay!
2015-01-20T21:20:41Z pjb: So we may (define (rule-operator rule)   (car (car (cdr rule))))
2015-01-20T21:20:59Z pjb: Try it: (let  ((rule '(--> (* int int) int)))   (rule-operator rule))
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2015-01-20T21:21:40Z lrs: pjb, Im not sure.. What should I try?
2015-01-20T21:21:59Z pjb: All the expression that I type here, you should evaluate them in a scheme REPL!
2015-01-20T21:22:13Z lrs: Yeah, I do. But the --> is strange
2015-01-20T21:22:29Z lrs: And (let  ((rule '(--> (* int int) int)))   (rule-operator rule)) doesnt work since rule-operator is not defined yet
2015-01-20T21:22:31Z pjb: We could use another format.
2015-01-20T21:22:55Z pjb: We could use the format: (when applying * on arguments of type (int int) we obtain a value of type int)
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2015-01-20T21:23:13Z lrs: Right
2015-01-20T21:23:14Z pjb: Then (define (rule-operator rule) (car (cdr (cdr rule))))
2015-01-20T21:23:36Z pjb: (rule-operator '(when applying * on arguments of type (int int) we obtain a value of type int))  returns: *
2015-01-20T21:23:58Z pjb: but it's simplier to write (--> (* int int) int)
2015-01-20T21:24:09Z lrs: Oh, I get it
2015-01-20T21:24:16Z lrs: But, we need to make it recursive
2015-01-20T21:24:20Z pjb: It's like (+ 2 3) which is written in maths 2 + 3.
2015-01-20T21:24:35Z pjb: Here (--> (* int int) int) would be written in maths (* int int) --> int
2015-01-20T21:24:42Z pjb: No we do not.
2015-01-20T21:24:49Z pjb: That's the point of using rules.
2015-01-20T21:24:57Z pjb: You don't need to make recursive rules.
2015-01-20T21:25:00Z lrs: Hmmm
2015-01-20T21:25:06Z pjb: The recusivity is managed by the check function itself.
2015-01-20T21:25:19Z lrs: So I dont have to make rule operator recursive so it always takes the first operator.. right?
2015-01-20T21:25:21Z lrs: What about
2015-01-20T21:25:26Z pjb: Yes.
2015-01-20T21:25:29Z lrs:  > (check '(= (< (+ int int) (quotient int int)) (> int int)))
2015-01-20T21:25:31Z pjb: We only consider one operator at a time.
2015-01-20T21:25:41Z lrs: WHat does rule operator do here.. Or lets say. I put rule-operator there instead.
2015-01-20T21:25:44Z pjb: Now, write the three rule-… function, and paste them.
2015-01-20T21:25:55Z lrs: Or, wait, it only checks one list at a a time? one cons of a pair?
2015-01-20T21:26:04Z pjb: one rule yes,.
2015-01-20T21:26:09Z lrs: Im not really sure about the structure of rule-operator
2015-01-20T21:26:17Z lrs: Or how it correlates to the other programs
2015-01-20T21:26:54Z pjb: it is called by our lambda predicate.
2015-01-20T21:27:00Z pjb: So we have to implement it.
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2015-01-20T21:29:50Z lrs: pjb, Why do you call it lambda?
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2015-01-20T21:30:04Z pjb: It's because it has no name.
2015-01-20T21:30:12Z lrs: I mean, why do yo ustart calling it lambda now? We had previous ones that have lambda in them right
2015-01-20T21:30:14Z lrs: Ah ok
2015-01-20T21:30:15Z pjb: lambda is the operator used to introduce unnamed functions.
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2015-01-20T21:30:39Z lrs: But when you write define, you have a lambda in it still am i not right?
2015-01-20T21:30:59Z pjb: Yes, but those functions are "named" by the variable to which we bind them.
2015-01-20T21:31:10Z lrs: YOu write (define(function x) or (define function (lambda (x))
2015-01-20T21:31:14Z lrs: And its the smae hting.
2015-01-20T21:31:15Z pjb: Yes.
2015-01-20T21:31:26Z lrs: Right. Ok. I just wanted to be sure so it wasnt something weird thing about predicate
2015-01-20T21:31:35Z pjb: No, nothing weird.
2015-01-20T21:32:17Z lrs: Ok
2015-01-20T21:32:27Z pjb: We could have named it, but since we use variable from the outer scope, the parameters operator and arg-types, this lambda predicate is actually a closure, and we would have to give it a name inside the apply-rules function.
2015-01-20T21:33:04Z pjb: We could have written: (let* ((match? (lambda (rule) (and …))) (matching-rule (find-if match? rule) )) (if …))
2015-01-20T21:33:58Z pjb: But actually, this init form for the matching-rule, we should probably extract it into its own function (matching-rule operator arg-types), because such a function could be useful to perform other operations than just apply-rule.
2015-01-20T21:34:23Z pjb: But for now, write the three rule-… functions and paste them in the annotation.
2015-01-20T21:34:53Z pjb: http://paste.lisp.org/display/145296
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2015-01-20T21:37:58Z lrs: pjb, Ok, give me a minute to think
2015-01-20T21:38:04Z pjb: sure.
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2015-01-20T22:10:55Z lrs: pjb, http://paste.lisp.org/display/145296#1 I suck so bad
2015-01-20T22:13:01Z pjb: Have you tested them?
2015-01-20T22:13:23Z lrs: Im not sure how to test them. I dont get your symbol thing...
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2015-01-20T22:13:44Z pjb: lrs: http://paste.lisp.org/+3440/2
2015-01-20T22:13:49Z pjb: use this definition for rules.
2015-01-20T22:14:00Z pjb: Then type in the REPL: (car rules) to find the first rule.
2015-01-20T22:14:23Z pjb: and type: (rule-argument-types (car rules))  to find the argument types of the first rule.  It should be (int int)
2015-01-20T22:14:59Z pjb: similarly, (rule-operator (car rules)) should return +
2015-01-20T22:15:56Z pjb: unfortunately, I can't find your annotation a rule-result-type definition (the parentheses don't balance).  The least thing is to ensure that parentheses balance.
2015-01-20T22:16:11Z pjb: You should use emacs with paredit-mode to help you ensure it.
2015-01-20T22:16:42Z pjb: I'm sorry that I don't know the features available in DrRacket.  Why does it let you write such incomplete expressions?
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2015-01-20T22:18:03Z pjb: I've added a test function: http://paste.lisp.org/display/145296#3
2015-01-20T22:18:10Z lrs: rules : undefined
2015-01-20T22:18:33Z pjb: You should define it. In paste #2, there's a definition.
2015-01-20T22:18:48Z pjb: copy and paste that text to your REPL and to your source file.
2015-01-20T22:18:55Z lrs: I did
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2015-01-20T22:20:41Z pjb: In emacs, I type M-x run-scheme RET to launch mit-scheme in an inferior scheme process, and then I just type scheme code in lrs.scm and then C-x C-s to save the file, and C-x C-l to load the file in the inferior scheme process (mit-scheme).  The results are printed in the *scheme* buffer.
2015-01-20T22:21:56Z pjb: I can try out expressions in the source buffer by typing C-x C-e after an expression (result printed in the *scheme* buffer), or I can type expressions followed by RET in the REPL (the *scheme* buffer).
2015-01-20T22:22:36Z pjb: So when I write a function, I actually write little subexpressions, and evaluate them to check them while I build the more complex expression.
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2015-01-20T22:23:00Z pjb: I may define test data or use let as I've shown above.
2015-01-20T22:23:21Z pjb: Probably something similar can be done in DrRacket.
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2015-01-20T22:28:35Z lrs: pjb, You mean that i should paste in the "terminal" ?
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2015-01-20T22:29:09Z lrs: pjb, Ah, that works
2015-01-20T22:29:17Z lrs: It gives me int
2015-01-20T22:29:58Z pjb: it's wrong the argument types should be a list of types (int int) in the case of the first rule.
2015-01-20T22:30:13Z pjb: And here I get (int), not int.
2015-01-20T22:30:28Z pjb: parentheses are very very very important in lisp.
2015-01-20T22:32:14Z lrs: I get int
2015-01-20T22:33:31Z pjb: Then your function is wrong. Correct it.
2015-01-20T22:33:42Z lrs: (car (cdr (cdr rule))))
2015-01-20T22:33:49Z lrs: Oh fk
2015-01-20T22:34:21Z LeoNerd: (caddr rule)   surely? ;)
2015-01-20T22:34:49Z pjb: (define (rule-argument-types rule) …) evaluate, try out, check result.
2015-01-20T22:35:36Z lrs: pjb, What does --> do to the cdr and whatever
2015-01-20T22:35:41Z lrs: It doesnt destroy it right
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2015-01-20T22:36:20Z pjb: --> doesn't do anything.  It's a symbol. Same as whatchamacall, thingy, schmiblick, + or blah.
2015-01-20T22:37:09Z lrs: pjb, What do you want the output to be?
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2015-01-20T22:37:54Z lrs: (car (cdr (cdr rule))))
2015-01-20T22:37:59Z lrs: (rule-operator (car rules))
2015-01-20T22:38:00Z lrs: = int
2015-01-20T22:38:47Z pjb: So it's wrong.
2015-01-20T22:38:57Z pjb: (rule-operator (car rules)) should be +
2015-01-20T22:39:05Z pjb: (rule-argument-types (car rules)) should be (int int)
2015-01-20T22:39:10Z pjb: (rule-result-type (car rules)) should be int
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2015-01-20T22:49:53Z pjb: You know what car and cdr do?
2015-01-20T22:49:57Z lrs: Yeah
2015-01-20T22:50:02Z lrs: Im just having trouble with argument-types
2015-01-20T22:50:05Z pjb: Why?
2015-01-20T22:50:39Z pjb: (rule-argument-types '(--> (+ int int) int)) = (int int).  What is (int int) in the rule?
2015-01-20T22:51:33Z pjb: It's the rest of the second element of that list.
2015-01-20T22:51:35Z lrs: (cdr (car rule))
2015-01-20T22:51:45Z pjb: rest is cdr, and second is cadr or (car (cdr .))
2015-01-20T22:51:46Z pjb: (cdr (car (cdr '(--> (+ int int) int)))) = (int int)
2015-01-20T22:52:03Z pjb: So (define (rule-argument-types rule) (cdr (car (cdr rule))))
2015-01-20T22:52:14Z pjb: It's not more complicated.
2015-01-20T22:53:26Z pjb: And similarly, rule-result-type is the third element of the list. so (car (cdr (cdr rule)))
2015-01-20T22:55:43Z lrs: > (rule-result-types (car rules))  int > (rule-argument-types (car rules))  (int int) > (rule-operator (car rules))  +
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2015-01-20T22:59:16Z pjb: good.  Now paste them, and start writing find-if.
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2015-01-20T23:01:19Z lrs: pjb, Ok, give me some time to think
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2015-01-20T23:14:07Z lrs: pjb, Could you tell me the purpose of find-if again?
2015-01-20T23:14:38Z pjb: return the first element in the list for which predicate? returns true, or else the symbol nil.
2015-01-20T23:14:38Z pjb:  
2015-01-20T23:15:44Z lrs: pjb, Dont I need to write out predicate too then
2015-01-20T23:16:09Z pjb: No. The predicate is a parameter.
2015-01-20T23:16:20Z pjb: (define (find-if predicate? list)  …)
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2015-01-20T23:35:06Z lrs: pjb, (define (find-if predicate lst)
2015-01-20T23:35:07Z lrs:   (if (predicate? lst))(car lst)
2015-01-20T23:35:07Z lrs:   'nil)
2015-01-20T23:35:10Z lrs: :S
2015-01-20T23:36:58Z pjb: return the first element in the list for which predicate? returns true, or else the symbol nil.
2015-01-20T23:37:12Z pjb: It should call predicate? on elements of the list. Not on the whole list.
2015-01-20T23:37:25Z pjb: And for that, the list must not be empty.
2015-01-20T23:37:45Z lrs: pjb, Right. So it needs a recursion thing
2015-01-20T23:37:48Z lrs: BUt I dont know how to write it
2015-01-20T23:37:53Z pjb: In general, when you process a list, the first thing you must do is to test whether the list is empty.
2015-01-20T23:38:02Z lrs: Yes
2015-01-20T23:38:07Z lrs: So (null? lst)
2015-01-20T23:38:48Z pjb: yes.
2015-01-20T23:39:29Z lrs: (define (find-if predicate lst)
2015-01-20T23:39:29Z lrs:   (if ((null? lst) 'nil)
2015-01-20T23:39:29Z lrs:   (if (predicate? lst))(car lst)
2015-01-20T23:39:34Z pjb: If you read from the start to the point where this problem was given to you, you should find explications about how to write a recursive function.
2015-01-20T23:39:39Z lrs: I know i need to fix the syntax somehow
2015-01-20T23:39:53Z pjb: You should use cond instead of if, since you will have several tests to check in sequence.
2015-01-20T23:39:55Z lrs: How do you mean from the start?
2015-01-20T23:40:38Z pjb: Because using car/cdr to find elements in a list, and writing a recursive procedure is explained in general between the start of the book/tutorial, and the point where they ask you to write procedures requiring it.
2015-01-20T23:41:10Z pjb: What book do you read? Where did you find this exercice?
2015-01-20T23:41:30Z lrs: This is an exercise from my course
2015-01-20T23:41:49Z pjb: So, what did your teacher tell you about recursive functions?
2015-01-20T23:42:14Z lrs: Well, I know how it work. BUt i forget the "standard way" of writing it in scheme
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2015-01-20T23:43:03Z lrs: pjb, find-if is a boolean right?
2015-01-20T23:43:10Z pjb: No.
2015-01-20T23:43:15Z pjb: It's a selector.
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2015-01-20T23:43:20Z lrs: No wait, forget what i sad
2015-01-20T23:43:22Z pjb: It selects an element from the list.
2015-01-20T23:43:38Z pjb: And unless the predicate returns true for #f, it will always return a true value.
2015-01-20T23:43:58Z lrs: (define (find-if predicate lst)
2015-01-20T23:43:59Z lrs:   (cond ((null? lst) ... )
2015-01-20T23:43:59Z lrs:      ((predicate? lst))(car lst)))
2015-01-20T23:44:13Z pjb: http://htdp.org/2003-09-26/Book/curriculum-Z-H-7.html#node_sec_4.3
2015-01-20T23:44:15Z pjb: http://htdp.org/2003-09-26/Book/curriculum-Z-H-13.html#node_idx_896
2015-01-20T23:44:34Z pjb: Don't call predicate on the list!
2015-01-20T23:44:43Z pjb: Only on _elements_  of the list.
2015-01-20T23:44:58Z lrs: (define (find-if predicate lst)
2015-01-20T23:44:58Z lrs:   (cond ((null? (car lst)) ... )
2015-01-20T23:44:58Z lrs:      ((predicate? lst))(car lst)))
2015-01-20T23:44:59Z pjb: And you must do something when predicate? returns false.
2015-01-20T23:45:01Z lrs: woops
2015-01-20T23:45:16Z iKlsR joined #scheme
2015-01-20T23:45:21Z lrs: (define (find-if predicate lst)
2015-01-20T23:45:22Z lrs:   (cond ((null? (car lst)) ... )
2015-01-20T23:45:22Z lrs:      ((predicate? (car lst))...)))
2015-01-20T23:45:23Z pjb: You should really study htdp.
2015-01-20T23:45:26Z lrs: Sorry if I spam too much
2015-01-20T23:45:51Z pjb: What do you return when the list is empty?
2015-01-20T23:46:00Z pjb: (find-if even? '()) --> ?
2015-01-20T23:46:29Z lrs: (car lst)
2015-01-20T23:47:08Z lrs: "return the first element in the list for which predicate? returns tru"
2015-01-20T23:47:09Z pjb: find-if returns the first element in the list for which predicate? returns true, or else the symbol nil.
2015-01-20T23:47:23Z pjb: If there's no element in the list, then we are in the "else" case.
2015-01-20T23:47:29Z pjb: else the symbol nil is returned.
2015-01-20T23:47:35Z pjb: So you should return the symbol nil.
2015-01-20T23:47:57Z lrs: (define (find-if predicate lst)
2015-01-20T23:47:57Z lrs:   (cond ((null? (car lst)) 'nil )
2015-01-20T23:47:57Z lrs:      ((predicate? (car lst))...)))
2015-01-20T23:48:07Z pjb: Notice that in scheme when lst is (),  (car lst) will signal an error.
2015-01-20T23:48:12Z pjb: Ok.
2015-01-20T23:48:25Z pjb: Now when (predicate? (car lst)) is true, what should be returned?
2015-01-20T23:48:40Z lrs: (car lst) ?
2015-01-20T23:48:50Z pjb: yes.
2015-01-20T23:48:59Z jlongster joined #scheme
2015-01-20T23:49:13Z lrs: (define (find-if predicate lst)
2015-01-20T23:49:13Z lrs:   (cond ((null? (car lst)) 'nil )
2015-01-20T23:49:13Z lrs:      ((predicate? (car lst))(car lst))))
2015-01-20T23:49:19Z pjb: and finally, if predicate? returns false, for the first element of the list , what should we return?
2015-01-20T23:49:52Z lrs: When predicate is false, that means there is no predicate soooo
2015-01-20T23:50:12Z pjb: No. that just means that (predicate? (car lst)) is false. But what about (cdr lst)?
2015-01-20T23:50:28Z lrs: oh right, thats where the recursion comes in
2015-01-20T23:50:33Z pjb: yes
2015-01-20T23:50:44Z lrs: (find-if (cdr lst))
2015-01-20T23:50:53Z pjb: takes two arguments.
2015-01-20T23:51:32Z pjb: (find-if predicate? (cdr lst))
2015-01-20T23:51:40Z lrs: Ah
2015-01-20T23:52:35Z lrs: (define (find-if predicate lst)
2015-01-20T23:52:36Z lrs:   (cond ((null? (car lst)) 'nil )
2015-01-20T23:52:36Z lrs:      ((predicate? (car lst))
2015-01-20T23:52:36Z lrs:       (find-if predicate? (car lst)))))
2015-01-20T23:53:19Z pjb: Check http://htdp.org/2003-09-26/Book/curriculum-Z-H-7.html#node_sec_4.3 or http://www.schemers.org/Documents/Standards/R5RS/HTML/r5rs-Z-H-7.html#%_idx_106 for the syntax of cond.
2015-01-20T23:53:19Z rudybot: http://tinyurl.com/3gogwm
2015-01-20T23:53:54Z jlongster quit (Ping timeout: 264 seconds)
2015-01-20T23:53:54Z pjb: And you seem to have some problem with your short term memory.
2015-01-20T23:54:42Z pjb: how ((predicate? (car lst))(car lst))  and  (find-if predicate? (cdr lst))  get merged into    (define (find-if predicate lst)  (cond ((null? (car lst)) 'nil )    ((predicate? (car lst))    (find-if predicate? (car lst))))) ?
2015-01-20T23:54:56Z pjb: Do you use emacs?
2015-01-20T23:55:21Z pjb: You should use emacs, to edit text and code, while IRCing.
2015-01-20T23:59:31Z lrs_: pjb, I have ADHD