Precendence vs short-circuiting in C

Question

int i=-3, j=2, k=0, m; 
m = ++i || ++j && ++k; 
printf("%d, %d, %d, %d\n", i, j, k, m);

Since ++ has more precedence than || and && in C, they are evaluated first and therefore the expression becomes m = -2 || 3 && 1. Now you can apply short circuiting but that produces incorrect answer. Why is that?

It's like a party in my eyes and only evil clowns are invited. — Jeff Watkins, Jun 14 '12 at 12:48
What is the output you are getting and What is output you are expecting? — Jay, Jun 14 '12 at 12:49
@ta.speot.is, you have a grammatical error in roflmoa (grammar nazi) — Shahbaz, Jun 14 '12 at 12:51

dan04 · Answer 1 · 2012-06-14T15:59:56.277

4

Precedence ≠ order of evaluation.

The short-circuiting behavior of || and && means that their left-hand sides are evaluated first, and

If the LHS of || evaluates to true (nonzero), the RHS is not evaluated (because the expression will be true no matter what the RHS is)
If the LHS of && evaluates to false (or zero), the RHS is not evaluated (because the expression will be false no matter what the RHS is)

In your example, ++i gets evaluated, and is equal to -2, which is nonzero, so the right-hand side of the || (that is, ++j && ++k) never gets evaluated: j and k are never incremented.

edited Jun 14 '12 at 15:59

answered Jun 14 '12 at 12:56

dan04

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If what we have is `!(++i || ++j && ++k)` instead, does `j` and `k` get incremented? – Pacerier Nov 21 '13 at 21:33

Shahbaz · Answer 2 · 2012-06-14T13:03:53.927

The ++s don't execute before the expression. Only ++i executes, which indicates that the result of the expression will be 1, therefore the rest of the expression is not evaluated (short circuit).

Your code is equivalent to:

if (++i)
    m = 1;
else
    if (!++j)
        m = 0;
    else if (!++i)
        m = 0;
    else
        m = 1;

This means that once ++i is evaluated to true, the else part is never executed.

Precendence vs short-circuiting in C

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