Precendence vs short-circuiting in C [closed]

Answer 1

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.

Answer 2

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.