#include<stdio.h>
#include<conio.h>
main()
{
int i=2;
i=i---i--;
printf("%d",i);
getch();
}
I don't understand why output of this program is 1. I guess it may be precedence of operators. but I don't figure out why this.
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Vishvendra Singh
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Modifying the same variable twice between sequence points is undefined. Don't do it. – Barmar Jan 06 '14 at 06:56
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Apart from the fairly obvious undefined behavior, `i=i---i--;` is tokenized, via the "maximal munch" rule, as `i = i -- - i -- ;`. – Keith Thompson Jan 06 '14 at 06:58
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conio and getch? Seriously? Upgrade your compiler! – paxdiablo Jan 06 '14 at 06:59
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What are your trying to achieve with this instruction (i=i---i--;). – UserM Jan 06 '14 at 07:00
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Whenever you see some weird structure like `i=i---i--;`, remember it doesn't do what you intuitively think it does. – Ankur Jan 06 '14 at 07:00
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@Barmar if it is undefined then why the program giving output and why the output is 1. – Vishvendra Singh Jan 06 '14 at 07:01
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"Undefined behavior" means the program can do anything. That includes giving output. – Barmar Jan 06 '14 at 07:01
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@paxdiablo which compiler do you ????? – Vishvendra Singh Jan 06 '14 at 07:02
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@Barmar that's wrong dude this program not showing different thing every time i run this but gives a definite answer. so if you know why this answer is then help me ? – Vishvendra Singh Jan 06 '14 at 07:04
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Why would it show different output? "undefined" doesn't mean that it changes, it just means that the language doesn't specify what should happen. – Barmar Jan 06 '14 at 07:06
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It's not likely to differ with a particular implementation, but it may give a different answer with different implementations. – Barmar Jan 06 '14 at 07:06
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@Barmar bro you say language doesn't specify why this output is coming. then what the rule which specify this output. – Vishvendra Singh Jan 06 '14 at 07:08
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There is no "rule", it's whatever the implementor happened to decide to do. – Barmar Jan 06 '14 at 07:11