Undefined Value 2 Revision as of 10:55, 12 December 2013 by Arjun Suresh (talk | contribs) (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

<syntaxhighlight lang="c">

1. include<stdio.h>

int main() {

  int a = 2;
  a = ++a / a++;
  printf("%d", a);
  
  return 0;

} </syntaxhighlight>

Solution

The answer to this is "Undefined". C standard says that the side effects of an operation (for b = ++a; modification of a is a side-effect) need to be completed only before the next sequence point. This relaxation is given so that the compiler would be able to generate the most optimal code (which run faster). But as a consequence, programmer shouldn't do a read and a write from a memory location within a sequence point or otherwise the result would be undefined.

So, in the statement a = ++a / a++;

There are two side effect on a, which can happen anytime before the next sequence point. So, as per C standard, the output is simply UNDEFINED.

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