00:00:15 I fail to see how. 00:00:15 (begin . definitions) is a different thing from (begin definition ... expression ...) 00:00:29 -!- Nshag [user@lns-bzn-37-82-253-43-245.adsl.proxad.net] has quit [Ping timeout: 245 seconds] 00:00:30 At the top level? 00:00:33 Yes. 00:00:48 I fail to see how. 00:01:17 Is the `foo' in (define-syntax bar (syntax-rules () (begin (define foo 1) (set! baz (lambda () foo))))) an introduced name? 00:01:19 Because (begin . defs) is the same as defs; the begin is not a scope. 00:01:39 -!- bweaver [~user@75-148-111-133-Chattanooga.hfc.comcastbusiness.net] has quit [Quit: ERC Version 5.3 (IRC client for Emacs)] 00:01:58 In that case, clearly yes; that is of the form (begin definition ... expression ...) 00:02:18 Is the `foo' in (define-syntax bar (syntax-rules () (begin (define foo 1) (define baz (lambda () foo))))) an introduced name? 00:02:20 define-record-type should just signal an error (possibly not until runtime) if any fields have the same symbol name. 00:02:22 Chicken is indeed buggy in that respsect. 00:04:49 (BTW, I'm assuming that "((bar) ...) is implied around all these begins. 00:05:25 In your most recent cae, foo and baz are not introduced names, I believe. 00:06:16 case, even 00:07:13 I don't understand what you mean by ((bar) ...) 00:08:01 Individual syntax rules are of the form (pattern expansion), but yours do not appear to be so: the (begin . body) is in the pattern. 00:08:10 chandler: Your syntax was wrong. 00:08:19 Sorry. Perils of trying to code at an IRC input line. 00:08:25 (define-syntax bar (syntax-rules () ((bar) (begin (define foo 1) (define baz (lambda () foo)))))) 00:08:30 Exactly. 00:08:36 I thought jcowan was referring to where the syntax was used, and was confused. 00:09:00 I'm not sure I understand your reasoning about when `define' acts as a binding form. 00:09:34 Suppose now that I write: (define-syntax bar (syntax-rules () ((_) (begin (define foo 1) (baz foo)))) . Is `foo' an introduced name? 00:09:38 I'm not telling you what `baz' is yet. 00:10:16 (By the way, Felix would dispute that Chicken is buggy in this regard; this behavior is by design.) 00:10:29 -!- wingo [~wingo@81.38.185.122] has quit [Ping timeout: 245 seconds] 00:12:48 On the theory that all top-level defines merely assign? 00:13:14 -!- fabe [~fabe@p54A7ECF6.dip.t-dialin.net] has quit [Remote host closed the connection] 00:13:34 Yes. 00:14:24 I could never decide if that was a bug or not myself. 00:14:51 Well, R5RS clearly permits either interpretation. 00:15:26 The R5RS is not infallible. It is wrong to permit Chicken's interpretation. 00:17:34 My own position is that the assignment behavior of `define' is utterly broken, and should be dispensed with. `define' should be consistently treated as a binding form in all contexts. `begin' should be consistently treated as an aggregator of bindings (some of them with unique names, in the case of expressions) in all contexts. This is the only behavior that provides any degree of sanity for macro authors. I'm not hopeful that I could ... 00:17:40 ... convince anyone else to adopt this view, however. 00:18:16 *jcowan* notes that in fact (define (y) (begin (define x 32) x)) is forbidden by R5RS, although allowed by Chicken. 00:18:52 And nobody could argue that the (define x 32) is not a binding form, not even Chicken. 00:19:56 Gambit and Chibi swallow that too, whereas Guile and scsh adhere to the R5RS here. 00:21:16 I couldn't think of a good reason to forbid that, and it just requires extra code to check for it. 00:21:43 I would actually prefer to permit that, as well as (let ((f (lambda (x) x))) (f (begin (define y 1) y))) . I would expect this to be controversial as well. 00:22:17 -!- pavelludiq [~quassel@91.139.194.233] has quit [Read error: Connection reset by peer] 00:23:58 Gambit will not swallow that, but Chicken and Chibi continue to be happy. 00:24:51 However, I do not think begin can be treated as nothing but an aggregator of bindings, because it has to return a value if there's one to return. 00:24:58 Really? I need to fix that. 00:26:05 *jcowan* also notes that R5RS doesn't explicitly license (begin . definitions) at the top level, only internally. 00:26:54 IOW, it is not all one whether you write (begin (define x 1) y) or (begin (define x 1) (define scooby-doo y)). 00:27:50 That's news to me. 00:27:55 "At the top level of a program (begin ...) is equivalent to the sequence of expressions, definitions, and syntax definitions that form the body of the begin." 00:28:00 At other than the top level, I mean. 00:28:16 I'm afraid you've confused me again. 00:28:49 No, I am the one who is confused. 00:28:55 Erase all that. 00:29:04 Regarding `begin' as an aggregator of bindings, I would in fact prefer it if that was the case - which then implies that ((lambda () (define x 1) (define y 2))) -> 2, I suppose. 00:29:29 Oh, if you are willing to swallow that, then quite so. 00:29:42 So begin does not merely aggregate bindings; it also returns the value of the last binding. 00:29:59 saccade_ [~saccade@209-6-54-113.c3-0.smr-ubr1.sbo-smr.ma.cable.rcn.com] has joined #scheme 00:30:15 Sgeo_ [~Sgeo@ool-18bf618a.dyn.optonline.net] has joined #scheme 00:30:15 Whatever receives the bindings does that. 00:30:19 In this case, it is `lambda'. 00:30:30 -!- Sgeo_ [~Sgeo@ool-18bf618a.dyn.optonline.net] has quit [Read error: Connection reset by peer] 00:31:02 However, lambda and begin don't have the same syntax: lambda takes a body, whereas begin does not. 00:31:28 What's your definition of "a body"? 00:31:55 "a (that is, the body of a lambda, let, let*, letrec, let-syntax, or letrec-syntax expression or that of a definition of an appropriate form)" 00:32:05 okay, that's technically recursive, but that's the point. 00:32:44 begin is defined as (begin expr1 expr2 ...); no definitions are allowed, *unless* it is a top-level-begin or a begin in the definition area of a body. 00:32:58 So there are three different begins with three different syntaxes. 00:33:49 -!- Sgeo [~Sgeo@ool-18bf618a.dyn.optonline.net] has quit [Ping timeout: 245 seconds] 00:34:00 Right. I dislike this situation immensely. `begin' should denote one concept, and the concept I am proposing is (I believe) a superset of what the R5RS describes as a , and close to what the R6RS concept of a program top level. 00:34:18 I'm sure that wasn't clear, now that I read it. 00:34:34 What `begin' should accept is a superset of what the R5RS describes as a . 00:34:52 In other words, if (begin 3 (define x 4) x) is allowed in one context, it should be allowed in all contexts. 00:35:13 What is the scope of x in that context? 00:35:18 s/context/expression 00:36:28 The scope is determined by whatever receives the bindings returned by `begin'. My internal mental model is of multiple syntactic return values, each denoting a binding. 00:36:38 I'll be back in a few minutes. 00:37:24 I mean, does (begin x (define x 1) (define y 2)) return 1, 2, or the value bound to x outside the begin-expression? 00:38:26 It returns 2. The value of the first expression in the `begin' is 1. 00:38:58 I believe that form should be the same as (begin (define x) (define x 1) (define y 2)) . 00:39:03 Thanks. I need to go home. 00:39:09 -!- jcowan [~jcowan@nat/google/x-jdxldretzulhporw] has quit [Quit: Leaving] 00:39:25 _JFT_ [~JFT@modemcable204.87-177-173.mc.videotron.ca] has joined #scheme 00:40:29 Obviously, this idea is quite radical, but I so far haven't found another way to ensure that `define' and `begin' always mean the same thing. 00:41:10 -!- phao [~phao@189.107.207.145] has quit [Quit: Leaving] 00:50:04 drwho [~drwho@c-71-225-11-30.hsd1.pa.comcast.net] has joined #scheme 00:52:08 -!- kniu [~kniu@HOHOHO.RES.CMU.EDU] has quit [Remote host closed the connection] 00:52:15 -!- _JFT_ [~JFT@modemcable204.87-177-173.mc.videotron.ca] has quit [Quit: _JFT_] 00:55:48 -!- fradgers- [~fradgers-@5ad4c06e.bb.sky.com] has left #scheme 00:59:54 bokr [~eduska@95.154.102.124] has joined #scheme 01:09:37 -!- alexsuraci_ [~alexsurac@32.170.97.189] has quit [Quit: alexsuraci_] 01:13:52 -!- Adamant [~Adamant@unaffiliated/adamant] has quit [Ping timeout: 256 seconds] 01:20:59 -!- offby1 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[~jon@c-98-202-82-46.hsd1.ut.comcast.net] has joined #scheme 02:56:36 -!- Owner_ [~ae66d3ce@gateway/web/freenode/x-zfvcdokgkkuqwbjx] has quit [Quit: Page closed] 02:57:25 Owner_ [~ae66d3ce@gateway/web/freenode/x-rsuyiflbrxhzkfcs] has joined #scheme 02:57:58 kniu [~kniu@HOHOHO.RES.CMU.EDU] has joined #scheme 03:11:58 -!- KatieHuber is now known as KatieHuber|away 03:13:34 -!- MrFahrenheit [~RageOfTho@users-55-240.vinet.ba] has quit [Ping timeout: 264 seconds] 03:19:38 Guvmint done stole an hour of sleep from me 03:26:03 bombshelter13b [~bombshelt@76-10-149-209.dsl.teksavvy.com] has joined #scheme 03:27:28 -!- xwl_ [~user@esprx01x.nokia.com] has quit [Remote host closed the connection] 03:28:15 xwl_ [~user@esprx01x.nokia.com] has joined #scheme 03:32:03 dey done did dat 03:32:05 ayup 03:36:10 -!- MichaelRaskin [~MichaelRa@195.91.224.225] has left #scheme 03:42:47 *Daemmerung* spits 03:42:47 also, s/Guvmint/Gubmint 03:42:49 foof [~user@FL1-118-110-11-244.osk.mesh.ad.jp] has joined #scheme 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[~drwho@c-71-225-11-30.hsd1.pa.comcast.net] has quit [Quit: A known mistake is better than an unknown truth.] 10:10:02 -!- jmcphers [~jmcphers@218.185.108.156] has quit [Remote host closed the connection] 10:13:09 -!- ironChicken [~richard@mx.slab.org] has quit [Quit: Terminated with extreme prejudice - dircproxy 1.0.5] 10:16:34 http://dpaste.com/172446//plain/ 10:16:38 could someone help me with that code? 10:16:45 rstandy [~rastandy@net-93-144-16-65.t2.dsl.vodafone.it] has joined #scheme 10:17:19 My issue is this: I have these functions 10:17:38 (first) and (second). They should get a quote, see if it's a pair 10:18:02 if it's not, return the quote, if it's, they should get the operation in that quote (operation) 10:18:10 and apply it to its first and second. 10:18:33 and they don't quite look at the entire quote 10:18:33 phao: funny, I was just playing with the same example yesterday 10:18:37 in first you do a car first before you even check if it is a pair 10:19:01 good point. 10:19:14 or I don't get how you hand it in 10:19:23 I assume it's (first '( 1 . 2)) 10:19:31 melba [~blee@unaffiliated/lazz0] has joined #scheme 10:20:31 you don't obey the first commandment :) Always ask null? As the first question in expressing 10:20:34 an function. 10:20:35 :) 10:20:41 :) 10:20:58 -!- rapacity [~prwg@li30-188.members.linode.com] has quit [Changing host] 10:20:58 rapacity [~prwg@unaffiliated/rapacity] has joined #scheme 10:21:33 the problem I'm having has nothing to do with that afaik. 10:21:46 I should get (make-sum 1 1) if the quote is (1 + 1) 10:21:50 and I'm getting (1 + +) 10:21:54 ops 10:21:56 I'm getting 10:22:01 (make-sum 1 '+) 10:22:27 well, you say (else (list a '+ b)))) 10:22:36 which does exactly that 10:22:49 well, the point is that 10:23:02 in a quote '(1 + 1) 10:23:09 the first is 1, second is 1. 10:23:17 and since the first is the car and the second is the caddr 10:23:24 then my program should be generating (make-sum 1 1) 10:23:30 and not (make-sum 1 '+) 10:23:31 ah ok 10:24:13 http://dpaste.com/172449//plain/ <= check for pairs in this one 10:24:24 still your definition of first makes no sense to me then 10:24:34 why? 10:24:57 ah wait 10:26:11 hmm 10:26:15 I'm just seeing something 10:26:22 I don't need to check for pairs as you said 10:26:38 I am lost, where do you expect to get that makesum line? 10:26:46 (operation exp) 10:26:58 should return the procedure that the quote refers to 10:27:02 if it's a sum, it returns make-sum 10:27:08 ... product, ... make-product 10:27:09 ... 10:30:02 why do you start with the difficult example when (first '(1 + 1)) won't work? 10:30:24 won't it? 10:30:34 eval it on paper 10:30:37 it does here. 10:30:41 it returns me 1. 10:31:42 and I did it on paper I think 3 or 4 times. 10:31:47 it looks correct. 10:33:28 I've changed it to something more readable imo 10:33:38 http://dpaste.com/172453//plain/ 10:33:42 maybe you prefer using this one 10:39:05 hmmmmmm 10:39:07 strange!!! 10:39:11 it works in mit-scheme 10:39:24 (note: I was using mzscheme) 10:41:05 can it be a bug in mzscheme? 10:43:16 alvatar [~alvatar@143.119.20.95.dynamic.jazztel.es] has joined #scheme 10:43:49 foof [~user@FL1-118-110-11-244.osk.mesh.ad.jp] has joined #scheme 10:47:02 well 10:47:07 I've reported a bug to mzscheme 10:47:13 I hope I'm not wrong :-) 10:47:48 what's your input, which output do you expect and what's mzscheme's output? 10:48:41 mzscheme's output is: ((3 + +) + +) 10:48:44 I expect: 7 10:48:57 input is: '(x + ((3 + (2 + 1)) + 1)) 10:49:04 I run (first e) 10:49:07 and e is input. 10:49:16 ops 10:49:17 ops 10:49:26 (second e) 10:49:26 hehe 10:51:57 be more precise. Where's the bug? 10:52:23 good question. 10:52:31 I only know it's giving an output it should give. 10:52:41 ops 10:52:47 it shouldn't* 10:53:34 and you expect people to debug your whole program to find a bug in a well tested scheme implementation? 10:53:43 -!- ribbs [~ribbs@p024062.doubleroute.jp] has quit [Quit: Leaving...] 10:53:54 that's unlikely the case 10:54:45 and.. ? 10:56:01 pmd [~user@mail2.siscog.pt] has joined #scheme 10:56:18 phao: Whittle it down to a tiny case which demonstrates the bug clearly 10:56:23 -!- pmd [~user@mail2.siscog.pt] has left #scheme 10:56:47 honestly... that is a considerable small case that the bug happens 10:56:51 Often in doing so you find out the exact procedure which is doing it wrong 10:57:02 Or a bug in your program, for that matter :) 10:57:07 yes 10:57:11 the bug may be in my program. 10:57:18 What is considerably small? 10:57:18 but then, I'd have to report a bug in mit-scheme 10:57:25 Your paste is a huge program 10:57:27 because it gives me the result I expect. 10:57:41 this input is considerable small '(x + ((3 + (2 + 1)) + 1)) 10:57:52 Sure, but the program is large 10:57:58 oh yes 10:58:01 It shouldn't be more than 10 lines 10:58:07 Preferably less 10:58:20 the fact is that its late now. I've already reported 10:58:22 If it really is a bug in mit scheme, then you should be able to demonstrate it with a smaller program 10:58:25 but the code that 10:58:30 That's okay, you can annotate your bugreport 10:58:31 I told them that is buggy is the one 10:58:35 The maintainers will appreciate it 10:58:38 related to the procedures (first) and (second) 10:58:44 they don't need to look at the rest of the file. 10:59:31 note that the bug is in mzscheme. 10:59:39 mit-scheme gives me the expected results 10:59:42 mzscheme does not. 10:59:45 But what is the bug? 10:59:55 can you point your finger at the s-expression that does the wrong thing? 10:59:56 Does it not apply operation correctly? 11:00:04 Does it not take the car correctly? 11:00:15 Does pair? say #t where it shouldn't? 11:00:22 C-Keen, with sureness? no. But I'm almost sure that it's the part of the procedures first and second 11:00:27 that is the recursive one 11:00:29 Really, you should be able to make it as short as that 11:00:46 phao: that's the first step. try to find the sexpression that does not do what you expect 11:00:50 You should really make it so simple to say: this core procedure goes wrong on this input 11:01:00 Not: here's my program, it breaks. Please help me 11:01:10 I didn't do any of those. 11:01:22 You did 11:01:28 How do you know? 11:01:35 Like you said, they only have to look at two procedures 11:01:35 you didn't look at the bug report. 11:01:46 Then you should've put only those procedures in the bugreport 11:01:52 why? 11:01:56 that makes even less sense 11:02:00 Idon't know where the bug is. 11:02:06 Because the rest is unrelated to the bug, according to you 11:02:08 Ah! 11:02:08 maybe it's a bug on the part of mzscheme 11:02:12 that 11:02:16 loads the file 11:02:21 If you don't know where the bug is, you should try to find out where it is 11:02:24 maybe it's doing something wrong because of other parts of the code. 11:02:34 I'm not going to read mzscheme's source to find a bug dude. 11:02:43 Prod and poke it until you find out where exactly it goes wrong 11:02:47 you should read your own code to find the bug 11:02:49 You don't have to look at mzscheme's source 11:02:57 But you should pinpoint where in your code it goes wrong 11:03:08 there is a line in my code 11:03:12 (display (second e)) 11:03:15 it's the only display line 11:03:18 and it outputs the wrong answer. 11:03:24 Okay, that's a start 11:03:25 the problem is in second 11:03:29 second has only 4 lines 11:03:33 ok, that is less than 10 lines. 11:03:37 satisfied? 11:03:38 Jafet [~Jafet@unaffiliated/jafet] has joined #scheme 11:03:46 Can you reproduce it by calling second on an example input? 11:03:52 tbh I think second has 5 lines. 11:03:56 If not, the bug is elsewhere 11:04:09 [08:44] Can you reproduce it by calling second on an example input? 11:04:11 I didn't understand this. 11:04:26 phao: open a REPL and type (second 'whatever) to trigger the bug 11:04:32 If you evaluate the definition of SECOND in mzscheme, and then type in (second ....), can you get the bug? 11:04:34 (after loading the file) 11:05:15 C-Keen: I'd skip loading the file and only paste the relevant definitions in the REPL to eliminate interaction with other parts in the code 11:05:22 (load "e2.58.ss") 11:05:22 ((3 * *) + +)> 11:05:22 (second e) 11:05:22 ((3 * *) + +) 11:05:49 phao: Ok, so what do you think it is doing here? 11:05:55 phao: What does e look like if you replace (display (second e)) with (display e) (Just to make sure you haven't somehow mutated e so that second gets confused)? 11:05:56 no clue. 11:06:01 Which path in your code does it take? 11:06:22 it gets the operation of the expression, then it gets its first and second part 11:06:27 after that apply the operation to those parts. 11:06:29 Does it? 11:06:30 the POINT is that 11:06:35 if I do it outside the procedure 11:06:37 Or does it follow the else branch of the IF? 11:06:37 it works. 11:06:39 it I do this 11:06:47 ... 11:06:51 ((operation (caddr e)) ...) 11:06:53 it gets me 7 11:06:55 ops 11:06:56 not 7 11:07:00 but the correct output. 11:07:04 phao: Check what branch of the if it is following 11:07:31 For example, (if (pair? s) 11:07:38 > (display e) 11:07:38 (x * ((3 * (2 + 1)) + 1)) 11:07:47 (begin (display "first branch") ((operation s) (first s) (second s))) 11:07:49 > e 11:07:50 (x * ((3 * (2 + 1)) + 1)) 11:07:54 (begin (display "second branch") s) 11:08:03 what is (begin) ? 11:08:20 It allows you to put multiple expressions in sequence in a place where you can only put one expression 11:10:06 What does it print? 11:10:36 hi! 11:10:39 hi alvatar 11:11:13 what are the real advantages of using a "record" against defining your own selector methods for a "type"? 11:11:23 It seems that gambit is dealing with them in a different way 11:11:31 alvatar: records introduce new disjoint types 11:11:48 This means you can distinguish between a record of type and, for example, a list 11:12:00 I see 11:12:01 wait.. 11:12:08 SICP uses a kind of "label" for that 11:12:10 Also, records are often more convenient 11:12:25 Yeah, but a label isn't really opaque 11:12:26 like you have normal lists, but one of the elements is a identifier of the "type" 11:12:29 It's still a list 11:12:33 of course 11:12:35 pair? and list? will answer #t 11:12:40 so records are a different type then? 11:12:43 Each record is really its own type 11:12:43 yes 11:12:46 nice 11:12:50 Down to the core 11:12:56 Instead of bolted on :) 11:13:05 I thought they were r6rs though 11:13:08 second:cond:else:exp:(x * ((3 * (2 + 1)) + 1)) 11:13:08 first:cond:else:exp:((3 * (2 + 1)) + 1) 11:13:08 ((3 * *) + +) 11:13:18 Nah, all schemes have their own record types 11:13:22 ok 11:13:25 http://dpaste.com/172458//plain/ 11:13:27 that is the new code 11:13:30 They're often nonportable, but there's also SRFI-9 11:13:49 phao: eek! 11:13:52 I thought that everything in scheme were lists, but there are also vectors and records, as I'm discovering them 11:13:58 You're not making it easier on yourself (or us, for that matter) 11:14:08 alvatar: There are good reasons for those 11:14:16 yes of course :) 11:14:27 sjamaan, why? 11:14:51 phao: For one, you're still pasting the entire library, forcing people to scroll down and make out the one interesting function 11:15:03 And also, you're changing the entire structure of the code 11:15:06 but what if the bug 11:15:08 is in mzscheme 11:15:14 and it's not in evaluating my specific function 11:15:18 You're not debugging what you had, but writing new code 11:15:19 what if the bug is being trigerred 11:15:23 by some other part of my code 11:15:31 Exactly! 11:15:31 because of some mechanism of mzscheme that doesn't read scheme right 11:15:37 "Exactly!" 11:15:44 That's why I want you to try it out _in isolation_ first 11:15:45 that is why I think they need the entire code. 11:15:50 Debugging is a process of elimination 11:15:52 phao: a lot of assumptions, what do you do to verify those 11:15:55 You have a hypothesis. Test it! 11:16:06 C-Keen, besides reading mzscheme's source, I don't know how to . 11:16:20 phao: Forget about that. Test your hypothesis 11:16:20 my hypothesis, now 11:16:21 is that 11:16:23 phao: by running tests against your own code 11:16:27 as soon as 11:16:31 (operation exp) 11:16:35 returns, make-sum is called 11:16:46 and (first exp) isn't evaluated 11:16:50 phao: That's not the same hypothesis you posed just a minute ago 11:16:56 sjamaan, ohh 11:16:58 yes yes 11:16:59 it is not. 11:17:00 |12:14| ( phao) what if the bug is being trigerred 11:17:00 |12:14| ( phao) by some other part of my code 11:17:03 ^ test that, first 11:17:08 I don't know how to . 11:17:09 Take baby steps 11:17:17 You can test that by ripping out all other stuff 11:17:32 Make a file containing only SECOND, and the definitions it depends upon 11:17:35 lol 11:17:37 See if it still reacts the same 11:17:43 hehe, yes. didn't think of that. 11:17:54 If it does, your hypothesis is incorrect and you can move on to form a new one 11:18:11 If it doesn't, well then we can try to find out _what_ part of your code is causing the difference 11:23:57 you were right 11:24:05 it seems that expression is being modified 11:24:12 ? 11:24:22 http://dpaste.com/172464//plain/ 11:24:52 my output is 11:24:53 pff, still quite a pile of code. But better! 11:25:00 It cannot be trimmed 11:25:04 all the procedures there 11:25:09 are necessary for first and second to run 11:25:12 It can, if you stub out some of the procedures that don't matter 11:25:12 my output is: 11:25:14 second:cond:else:exp:(x * ((3 * (2 + 1)) + 1)) 11:25:15 operation:sum:exp:((3 * (2 + 1)) + 1) 11:25:15 first:cond:else:exp:((3 * (2 + 1)) + 1) 11:25:15 operation:product:exp:(3 * (2 + 1))((3 * *) + +) 11:25:15 operation:sum:exp:((3 * (2 + 1)) + 1) 11:25:15 # 11:25:21 sjamaan, they all matter. 11:25:36 this one: operation:product:exp:(3 * (2 + 1))((3 * *) + +) 11:25:49 phao: ok. do you think that operation in this case looks right? 11:26:00 even the first one? 11:26:33 the procude operation? yes, it does look right. 11:26:41 it seems that it's receiving the modified expression 11:26:42 from first. 11:26:46 -!- virl [~virl__@chello062178085149.1.12.vie.surfer.at] has quit [Remote host closed the connection] 11:26:51 so, maybe it's first that is modifying the damn thing. 11:27:26 phao: Try to reduce the input expression. Can you make it shorter? 11:27:48 Try to find exactly under what circumstances the bug is triggered 11:27:52 I can, wait.. 11:27:58 sjamaan, I think i know that 11:28:08 the bug is triggered when I pass a compound expression 11:28:09 Only when you have nesting of one deep, two deep? Or does the length of the list matter? 11:28:14 (1 + 1) works 11:28:17 Good 11:28:20 ((1 + 1) + 2) doesnt 11:28:24 Excellent! 11:28:25 let me do it 11:29:29 http://dpaste.com/172466//plain/ -- this also prints the value of s and f in second and first. 11:29:34 second:cond:else:exp:(3 * (2 + 1)):s:(2 + 1) 11:29:34 operation:sum:exp:(2 + 1) 11:29:34 first:cond:2nd:exp:(2 + 1) 11:29:34 (2 + +) 11:30:03 I'd switch to the simplified example 11:30:08 ((1 + 1) + 2) 11:30:13 It's less complicated 11:30:22 Complexity is your enemy here 11:30:34 The simpler expression will help you debug it easier 11:30:45 (2 + 1) 11:30:47 that is pair. 11:30:48 sure? 11:30:51 I mean 11:30:55 '(2 + 1) 11:31:02 try it in the repl: (pai? '(1 + 2)) 11:31:07 try it in the repl: (pair? '(1 + 2)) 11:31:18