C - 等式的评估顺序

Question

I have done a ton of research as to how the order of evaluation goes - but cannot figure out how it would go for this equation:

z = !x + y * z / 4 % 2 - 1

My best guess is (from left to right):

z = !x + {[([y * z] / 4) % 2] - 1}

Answer 1

Order of evaluation and operator precedence are two different things.

Your best guess is correct. All the multiplicative operators * / % have the same precedence, and bind left-to-right. The additive operator - has lower precedence. The unary ! operator binds more tightly than the multiplicative or additive operators. And the assignment operator = has very low precedence (but still higher than the comma operator).

So this:

z = !x + y * z / 4 % 2 - 1

is equivalent to this:

z = (!x) + (((y * z) / 4) % 2) - 1

But the operands may legally be evaluated in any order (except for certain operators like && , || , , , which impose left-to-right evaluation). If the operands are simple variables, this probably doesn't matter, but in something like:

z = func(x) * func(y);

the two function calls may occur in either order.

Answer 2

If you can't understand it, rewrite your expression

z = !x + y * z / 4 % 2 - 1

notx = !x;         /* you can move this line 1, 2, or 3 lines down */
tmp1 = y * z;
tmp2 = tmp1 / 4;
tmp3 = tmp2 % 2;
tmp4 = notx + tmp3;
tmp5 = tmp4 - 1;