Undefined Value 2 Revision as of 16:12, 18 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 multiple writes to a memory location within a sequence point or otherwise the result would be undefined.

Undefined behavior can occur if "Between two sequence points, an object is modified more than once,
or is modified and the prior value is read other than to determine the value to be stored"

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.

Undefined value means compiler can give any value. i.e.; different compilers or even different versions of the same compiler can give different answers. As per C standard, no programmer should write this code.

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