Vii)7. main(){ int i=-1,j=-1,k=0,l=2,m; m=i++&&j++&&k++||l++; printf("%d %d %d %d %d",i,j,k,l,m);}

0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

Questions by Beena   answers by Beena

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abhimanipal

  • Jan 30th, 2010
 

Although the answer posted above is correct. There is one mistake in the explanation. Beena says that && operator has higher precedence tha || operator so && is evaluated first. But interestingly in the expressions invloving &&, ||, ? the order of evaluation is from left to right. To prove my point here is an example

  1. main()
  2. {
  3.       int i=4,j=7;
  4.       j = j || i++ && printf("YOU CAN");
  5.       printf("%d %d", i, j);
  6. }
The answer to this question is 4 1
Here if we go from left to right we still get the same answer. But still something to think about :)

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