00:42:24 <Asztal> I'd rather .NET than perl ;)
00:42:42 <oerjan> i am obliged to warn you that this whole channel is in fact .NET too.
00:42:43 <Asztal> I'd rather CLR than Perl
00:43:05 <Razor-X> Think carefully my friend. CLR is horribly bloated. Perl is not.
00:43:18 <Asztal> and Perl is horribly fugly.
00:43:54 <Razor-X> C isn't horribly bloated. It's horribly fugly.
00:43:57 <Asztal> There's plenty of languages other than C on the CLR, you know :)
00:44:08 <Razor-X> Scheme beats 'em all, and then some.
00:44:24 <Asztal> actually, I'd have to say I prefer ruby, but I don't think there's a .rb ccTLD :(
00:44:42 <Asztal> C-like languages, sorry ;)
00:45:37 <Razor-X> Where's Scheme?! Bah! Where's Forth?!
00:45:51 <Asztal> You know you like COBOL.NET
00:46:07 * Asztal wonders if you can use it with managed Direct X
00:46:14 <Razor-X> I don't have Mono on this box.
00:46:28 <Razor-X> Talk about commercial and nonportable.
00:47:12 <Asztal> I suppose that's true. But somehow I still prefer it to OpenGL ;)
00:47:13 <oerjan> there is F#, which is an Ocaml clone
00:47:30 <lament> activex is somewhat of a non-issue
00:47:37 <lament> it doesn't even run in my OS
00:47:44 <lament> i don't have to care about museum software
00:48:19 <Razor-X> Asztal: Then your code stays on your platform.
00:48:46 <Razor-X> While Mac OS X and *Nix can share.
00:49:14 <Razor-X> OCaML's syntax looks god-awful.
00:49:31 <Asztal> But I tend to (try to) write them so they're more or less API-independant
00:49:32 <Razor-X> You would think S-expressions look bad, but OCaML is just *shudder*
00:50:21 <Razor-X> I'm no game programmer. Although I've been feeling like playing around with OpenGL in Scheme some time for the heck of it.
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08:43:47 <Razor-X> An ever shorter Forth quine:
08:45:47 <GreaseMonkey> damn, can't send blanks with a /raw command either :(
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12:07:44 <ka_1900> can any one plz tell me what is going on with this code !?
12:07:46 <ka_1900> http://pfo.pastebin.com/831922
12:08:13 <ka_1900> it is supposed to print a number ,.bt it prints another totally different number
12:29:02 <Asztal> that's undefined behaviour
12:29:39 <Asztal> the compiler is free to do whatever it likes
12:30:28 <ka_1900> no there is a logic behind this ,.
12:31:45 <Asztal> http://en.wikipedia.org/wiki/Sequence_point
12:32:37 <Asztal> you have to remember that function arguments can be evaluated in any order
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16:05:09 <ka_1900> this logic i sent u b4 is making me crazy
16:15:03 <RodgerTheGreat> there's an easy way to fix it- don't do multiple pre- and post- increments on the same variable in a function call
16:15:37 <RodgerTheGreat> just replace ++x with (x+1) and so on- then your code will function how you expect it to.
16:16:46 <ka_1900> the problem is nt in the wrong answer ,. the problem i have to solve ,. is why it is doing so ,. btw this is the right code
16:16:50 <ka_1900> http://pastebin.com/831918
16:20:14 <RodgerTheGreat> it's like Asztal said- function arguments can be evaluated in any order, and several of your arguments change the value of X.
16:21:01 <ka_1900> can we make it step by step together ???
16:22:06 <RodgerTheGreat> is nothing to go through step-by-step- you can effectively assume that the arguments you pass to your function are evaluated randomly, because it's undefined behavior
16:22:30 <ka_1900> int x = 5 ; printf("%d%d%d%d%d", x++, ++x, x, ++x, x++) ;
16:22:48 <ka_1900> the first printf should print 5
16:23:41 <ka_1900> and the second and third printf should print 7 ? right ?
16:23:46 <RodgerTheGreat> function arguments aren't necessarily evaluated from left to right, that's what I'm trying to get through to you.
16:23:58 <ka_1900> and the last ++x and x++ should print 8 ,
16:24:29 <ka_1900> but the calculation is done left to right and printed right to left ,. why ?!
16:24:48 <ka_1900> i mean the calculation gives 57788
16:25:50 <jix> but the calculation is done left to right and printed right to left ,. why ?! << it isn't done from left to right
16:26:21 <RodgerTheGreat> it's obviously being done right to left in this case, and it'd probably do something else on a different compiler.
16:27:11 <ka_1900> no there is nothing called undefiend behaviour ,. our proffessor told us it has a logic and we have to find out why
16:27:32 <jix> than read the c standard
16:28:26 <jix> and professors can be wrong
16:29:14 <RodgerTheGreat> whenever a function call is made with arguments that are expressions (x+5, x--, etc) they have to be evaluated before the actual function operates on these values (at an abstract level). The order in which arguments are evaluated is compiler-defined.
16:30:09 <RodgerTheGreat> the actual time at which the statements are evaluated in machinecode is determined by the compiler, which is *why* it's compiler-defined
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16:31:21 <ka_1900> so the answer that this is undefined behavioud
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20:06:19 <jix> i have a very cool idea for a new esolang!
20:06:50 <jix> nah doesn't work
20:07:01 <jix> i was lying i just wanted to test a script and needed people to talk
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20:51:15 <fizzie> You can't assume that arguments are evaluated randomly, because it's undefined behaviour; you must assume that _anything_ can happen, including the computer exploding.
20:51:25 <fizzie> It doesn't necessary evaluate any arguments at all.
20:52:06 <fizzie> (To answer the conversation that was here over four hours ago.)
20:52:59 <fizzie> (It's undefined behaviour since it modifies x multiple times between sequence points.)
20:56:36 <fizzie> Actual evaluation order is implementation-defined, but that's another issue. printf("%d%d\n", printf("foo\n"), printf("bar\n")); does not have any undefined behaviour, but what it prints is implementation-defined. It must be "foo\nbar\n44\n" or "bar\nfoo\n44\n", though. (Barring any IO errors or such.)
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21:25:49 <RodgerTheGreat> fizzie: yeah, true. I was just trying to explain things simply
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