Question

I have a code below and wanted to know what could be output.What i would like to see how different compiler would interpret this particular piece of code.

int main()
{
int i = -1, j = 2, k = 0, m;
m = ++i || ++j && ++k;
printf("\n %d %d %d %d \n", i, j, k, m);
return 0;
}

Now the question is while handling this code would compiler go by the rule which says

This expresion can be seen as

 m = ++i || (++j && ++k); 

as && has higher precedence over || and the result would be 0 2 0 1

or

This expresion can be seen as

m = ++i || (++j && ++k);

but compiler still will try to short circuit. so it evaluates ++i, since it’s 1,(++j&&++k) >are not evaluated.so ans is 0 3 1 1

Answer 1

The expression ++i || ++j && ++k will group like so due to precedence:

++i || (++j && ++k)

One of the few rules C has about evaluation order of expressions is that the left-hand side of the || and the `&& operators must execute before the right-hand side (for short circuiting). This is a non-optional language requirement.

That means that ++i has to be evaluated first. It will return 0 (it does not evaluate to 1 as you mistakenly mention in the question). Since that’s not enough to short circuit the || operation, the right side will need to be evaluated. A similar process will occur for evaluation the && operator (and both sides of the && operation will need to be evaluated).

The resulting output will be:

0 3 1 1 

regardless of the compiler. There is no undefined, unspecified, or implementation defined behavior evaluating that expression.

The expression ++i || ++j && ++k will group like so due to precedence:

++i || (++j && ++k)

One of the few rules C has about evaluation order of expressions is that the left-hand side of the || and the `&& operators must execute before the right-hand side (for short circuiting). This is a non-optional language requirement.

That means that ++i has to be evaluated first. It will return 0. Since that’s not enough to short circuit the || operation, the right side will need to be evaluated. A similar process will occur for evaluation the && operator (and both sides of the && operation will need to be evaluated).

The resulting output will be:

0 3 1 1 

regardless of the compiler. There is no undefined, unspecified, or implementation defined behavior evaluating that expression.

---

One way to get the output you mention in your question is to use the following expression instead:

m = --i || ++j && ++k;

In this case, --i evaluates to -2 and the short circuiting rule requires that the right side of the || operator not be evaluated. So j and k are left alone.