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2014-10-10T00:34:26Z pjb: lrs: don't despair, you can still learn and make progress! :-)
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2014-10-10T01:03:15Z _tca: adhd is the best i love it
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2014-10-10T01:45:03Z pjb: ADHD is not a learning disability.
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2014-10-10T02:11:48Z kvda: I imagine it wouldn't help with learning...
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2014-10-10T06:08:38Z sigjuice: SICP exercise 3.38 asks what different values can 'balance' end up with if these three operations execute concurrently. balance += 10, balance -= 20, and balance -= balance/2
2014-10-10T06:10:13Z sigjuice: I have a test program that sometimes produces 25.  Any ideas on how 25 is possible?
2014-10-10T06:13:59Z sigjuice: https://gist.github.com/khirbat/9ee7004f10ee0884ac2d
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2014-10-10T06:18:08Z fizzie: sigjuice: As 80-110/2. Have both Peter and Paul read the initial value of 'balance' (so they do (+ 100 10) and (- 100 20), respectively) and then have Paul first set balance to 80, arrange for Mary to evaluate one 'balance' so she has (- 80 (/ balance 2)), have Peter set balance to 110, and end up with (- 80 (/ 110 2)).
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2014-10-10T06:19:40Z fizzie: (Writing it as "balance += 10" and "balance -= 20" instead of "balance = balance + 10" and "balance = balance - 20" makes the operations look overly atomic.)
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2014-10-10T06:22:48Z sigjuice: is considering each operation as two operations still overly atomic? e.g. balance += 10 is "1. read balance 2. set balance to balance - 10"
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2014-10-10T06:24:32Z fizzie: That would depend on your concurrency model, really. But at least breaking it down to those two operations (you could use a different name in 2. -- e.g. "1. read balance as balance'. 2. set balance to balance' - 10" or something) is enough to explain the 25.
2014-10-10T06:29:07Z fizzie: (As long as Mary's case is written "1. Read balance as balance1. 2. Read balance as balance2. 3. Set balance to balance1 - balance2/2." with the caveat that steps 1 and 2 can occur in either order.)
2014-10-10T06:33:39Z sigjuice: thanks fizzie. I was working this out on paper as an interleaving of 3 threads where each thread has two operations (read balance and set balance).
2014-10-10T06:34:12Z sigjuice: and I failed to see that Mary's thread is actually three operations
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2014-10-10T06:41:27Z sigjuice: Is this how many interleavings are possible (2 + 2 + 3)!/(2! * 2! * 3!) ? Assuming Peter and Paul are two operations each and Mary is three operations.
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2014-10-10T06:48:52Z pjb: Hence the complexity of multithreaded programs.
2014-10-10T06:52:24Z sigjuice: Of course.  That is exactly what SICP says in the next section.
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2014-10-10T10:27:21Z vanila: what are the methods to determine when you can stack allocate a closures environment after CPS?
2014-10-10T10:27:36Z vanila: 0-CFA seems extremely difficult
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2014-10-10T13:24:48Z vanila: Hi
2014-10-10T13:24:53Z vanila: Repeating my question from earlier: what are the methods to determine when you can stack allocate a closures environment after CPS? 0-CFA seems extremely difficult
2014-10-10T13:33:44Z pjb: vanila: I don't know in particular with CPS. In general, you perform an escape analysis.
2014-10-10T13:34:13Z pjb: I don't know more specifics.
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2014-10-10T16:59:51Z vanila: http://research.microsoft.com/en-us/um/people/akenn/sml/CompilingWithContinuationsContinued.pdf
2014-10-10T16:59:52Z rudybot: http://tinyurl.com/p9dhqtr
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2014-10-10T19:37:06Z lrs: Im trying to create a procedure
2014-10-10T19:37:14Z lrs: (define f (cubic a b c))
2014-10-10T19:37:26Z lrs: That represents f(x)=x^3+ax^2+bx+c
2014-10-10T19:37:35Z lrs: ANyone have any ideas
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2014-10-10T19:43:02Z vanila: (define (f a b c)
2014-10-10T19:43:04Z vanila:  (lambda (x)
2014-10-10T19:43:08Z vanila:   ...))
2014-10-10T19:43:12Z vanila: You can fill in the ...
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2014-10-10T19:45:50Z lrs: vanila, THats what I have problem with
2014-10-10T19:45:51Z lrs: :S
2014-10-10T19:45:57Z _tca: but thats the easy part
2014-10-10T19:46:09Z _tca: just turn the math into prefix
2014-10-10T19:46:20Z lrs: Hmm
2014-10-10T19:46:21Z vanila: x^3+ax^2+bx+c is (+ (* x x x) (* a x x) (* b x) c)
2014-10-10T19:51:46Z lrs: define f (cubic a b c)
2014-10-10T19:51:50Z lrs: You cnat write it like that right
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2014-10-10T19:55:24Z vanila: in lisp you start with an open bracket
2014-10-10T19:55:28Z vanila: (define
2014-10-10T19:55:31Z vanila: rather than
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2014-10-10T19:55:32Z vanila: define
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2014-10-10T20:03:05Z lrs: Weird
2014-10-10T20:03:12Z vanila: What?
2014-10-10T20:03:21Z lrs: In the instructions it says (define f (cubic a b c)
2014-10-10T20:03:26Z lrs: Its wrong right
2014-10-10T20:03:28Z vanila: Did you understand my advice
2014-10-10T20:03:48Z lrs: ?
2014-10-10T20:03:53Z lrs: YEs
2014-10-10T20:03:56Z vanila: good
2014-10-10T20:04:22Z vanila: (define (cubic a b c)
2014-10-10T20:04:26Z vanila:  (lambda (x)
2014-10-10T20:04:30Z vanila:   ...))
2014-10-10T20:04:44Z vanila: If you do this you can define a function called 'cubic'
2014-10-10T20:04:48Z vanila: later you can use it
2014-10-10T20:04:53Z vanila: (define f (cubic a b c))
2014-10-10T20:04:57Z vanila: (f 3)
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2014-10-10T20:05:17Z lrs: I cant
2014-10-10T20:05:22Z lrs: (define f (cubic a b c))
2014-10-10T20:05:22Z lrs: (lambda (x)
2014-10-10T20:05:25Z lrs: Doesn work
2014-10-10T20:06:42Z vanila: Have you defined cubic?
2014-10-10T20:07:27Z lrs: (define f (cubic a b c))
2014-10-10T20:07:27Z lrs: (lambda (x)
2014-10-10T20:07:27Z lrs: (+ (* x x x) (* a x x) (* b x) c))
2014-10-10T20:07:32Z lrs: I cant add an last ) at the end
2014-10-10T20:07:34Z lrs: For some reason
2014-10-10T20:07:36Z _tca: you misread lrs
2014-10-10T20:07:37Z vanila: <vanila> (define (cubic a b c)
2014-10-10T20:07:37Z vanila: <vanila>  (lambda (x)
2014-10-10T20:07:37Z vanila: <vanila>   ...))
2014-10-10T20:07:41Z vanila: define cubic this awy
2014-10-10T20:07:49Z _tca: notice there is no f
2014-10-10T20:07:52Z lrs: Yeah
2014-10-10T20:07:55Z lrs: Thats why I said
2014-10-10T20:08:00Z lrs: <lrs> In the instructions it says (define f (cubic a b c
2014-10-10T20:08:08Z lrs: <lrs> Its wrong right
2014-10-10T20:08:09Z _tca: you are misunderstanding
2014-10-10T20:08:17Z _tca: that is for when you use it
2014-10-10T20:08:19Z _tca: this is for defining it
2014-10-10T20:08:31Z _tca: first you define cubic
2014-10-10T20:08:36Z _tca: then you define f using cubic
2014-10-10T20:08:40Z lrs: Oh
2014-10-10T20:10:45Z _tca: (define (cubic a b c) ...) (define f (cubic n1 n2 n3)) (f n4)
2014-10-10T20:12:42Z lrs: + )
2014-10-10T20:13:37Z lrs: Now I did > (define (cubic a b c) (f c))
2014-10-10T20:13:55Z vanila: no
2014-10-10T20:14:01Z vanila: 0 <vanila> (define (cubic a b c)
2014-10-10T20:14:02Z vanila: <vanila> <vanila>  (lambda (x)
2014-10-10T20:14:02Z vanila: <vanila> <vanila>   ...))
2014-10-10T20:14:10Z lrs:  (+ (* x x x) (* a x x) (* b x) c))
2014-10-10T20:14:19Z lrs: Thats what I got in the terminal window
2014-10-10T20:14:23Z lrs: (define (cubic a b c)
2014-10-10T20:14:23Z lrs: (lambda (x)
2014-10-10T20:14:23Z lrs: (+ (* x x x) (* a x x) (* b x) c)))
2014-10-10T20:14:31Z lrs: And input
2014-10-10T20:14:31Z lrs: > (define (cubic a b c) (f c))
2014-10-10T20:14:33Z vanila: tthat looks good
2014-10-10T20:14:35Z vanila: that does not
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2014-10-10T20:15:26Z lrs: hm
2014-10-10T20:15:32Z lrs: :S
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2014-10-10T20:18:48Z lrs: vanila, Whats the correct way?
2014-10-10T20:18:53Z _tca: lrs: take the last line i said and replace ... with (+ (* x x x) (* a x x) (* b x) c)
2014-10-10T20:19:06Z _tca: ignore that
2014-10-10T20:19:12Z vanila: <lrs> (define (cubic a b c)
2014-10-10T20:19:12Z vanila: <lrs> (lambda (x)
2014-10-10T20:19:12Z vanila: <lrs> (+ (* x x x) (* a x x) (* b x) c)))
2014-10-10T20:19:14Z vanila: this is correct
2014-10-10T20:19:25Z _tca: replace it with (lambda (x) (+ (* x x x) (* a x x) (* b x) c))
2014-10-10T20:19:46Z lrs: Yeah, I got that covered
2014-10-10T20:19:51Z lrs: Im wondering about the input
2014-10-10T20:19:53Z lrs: How do I call it
2014-10-10T20:20:19Z _tca:  (define f (cubic n1 n2 n3))
2014-10-10T20:20:26Z _tca: n1 n2 and n3 are the numbers of your choice
2014-10-10T20:20:43Z lrs: f undefined
2014-10-10T20:20:44Z lrs: ;_;
2014-10-10T20:21:00Z _tca: yes you have to define f with that line i just gave you
2014-10-10T20:21:11Z lrs: .
2014-10-10T20:21:14Z _tca: if you paste it directly in it will tell you n1 n2 and n3 are undefined
2014-10-10T20:21:17Z lrs: (define f (cubic 1 2 3))
2014-10-10T20:21:20Z _tca: there you go
2014-10-10T20:21:25Z _tca: now f is defined
2014-10-10T20:21:28Z _tca: now you can call (f 1)
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2014-10-10T20:21:37Z lrs: Nope
2014-10-10T20:21:40Z lrs: ;_;
2014-10-10T20:22:00Z lrs: f: undefined;
2014-10-10T20:22:34Z _tca: rudybot: (define (cubic a b c) (lambda (x) (+ (* x x x) (* a x x) (* b x) c)))
2014-10-10T20:22:34Z rudybot: _tca: Done.
2014-10-10T20:22:41Z _tca: rudybot: (define f (cubic 1 2 3))
2014-10-10T20:22:41Z rudybot: _tca: Done.
2014-10-10T20:22:46Z _tca: rudybot: (f 1)
2014-10-10T20:22:46Z rudybot: _tca: ; Value: 7
2014-10-10T20:22:56Z _tca: :)
2014-10-10T20:23:30Z lrs: Ugh
2014-10-10T20:23:34Z lrs: Forgot to run I think
2014-10-10T20:23:42Z lrs: Cool
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2014-10-10T20:24:46Z lrs: BTW
2014-10-10T20:24:54Z lrs: If you guys remember the next-highest function I wrote
2014-10-10T20:25:03Z lrs: Apparently it doesnt work if you have pairs of the same number..!
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2014-10-10T20:33:38Z CaptainRant: Why does hash-table-for-each take the table as a first arg, while normal for-each takes a list as the last arg ?
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2014-10-10T20:35:59Z taylanub: CaptainRant: no Scheme standard defines `hash-table-for-each'
2014-10-10T20:36:07Z CaptainRant: Oh
2014-10-10T20:37:25Z taylanub: other than that, I would actually ask the question why map and for-each take the procedure argument first.  when it's a lambda expression, it tends to be annoying to have the list argument dangling at the end...
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2014-10-10T20:38:00Z turbofail: well for map and for-each it's because you might pass in several lists
2014-10-10T20:38:15Z turbofail: and it'd be awkward to have a list, then a lambda, then more lists
2014-10-10T20:38:17Z taylanub: oh, good point
2014-10-10T20:38:34Z turbofail: although you could in theory make it take the function argument last
2014-10-10T20:38:38Z taylanub: well it could just take the last arg as a procedure...
2014-10-10T20:38:43Z mario-goulart: CaptainRant: I suspect it is to be consistent with the rest of the SRFI-69 API
2014-10-10T20:38:58Z CaptainRant: I never grasped this. I hoped this is the "scheme way" which should enlight me but, nah
2014-10-10T20:39:00Z mario-goulart: All them expect the hash-table as the first argument.
2014-10-10T20:39:09Z CaptainRant: ok, thanks :D
2014-10-10T20:39:31Z taylanub: CaptainRant: Scheme tries to be clean and consistent even on minor things, but it's difficult not to ever make such small mistakes...
2014-10-10T20:41:21Z taylanub: you'll find some small warts here and there even in Scheme APIs.  on the other hand, Scheme can probably be proud to tend to not bikeshed very much.  other languages will flame over superficial syntax and comment syntax, whereas Lisps just go with sexprs and be done with it :P
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2014-10-10T21:01:34Z asumu: Scheme just bikesheds about what module/macro system to use for 20 years.
2014-10-10T21:02:05Z vanila: answer: R7RS library system
2014-10-10T21:02:14Z vanila: it took only 20 years!
2014-10-10T21:02:56Z offby1: I still wish they'd painted it tan.
2014-10-10T21:02:58Z asumu: Yes, painting the bike shed a new color after choosing a color several years earlier...
2014-10-10T21:04:50Z vanila: asumu, what happened several years earlier?
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2014-10-10T21:08:04Z vanila: I don't know what you refer to
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2014-10-10T21:37:17Z taylanub: R6RS
2014-10-10T21:37:52Z taylanub: the R7RS-small library syntax is surprisingly more useful though
2014-10-10T21:38:26Z taylanub: and I've seen jcowan's proposal for the -large library syntax and it basically seals the deal IMO
2014-10-10T21:38:58Z taylanub: (of course, it's a strict extension of the -small library syntax)
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