使用size_t的负值切换

Question

If I have the following snippet:

size_t num = -1;

switch(num) {
    case -1:
        printf("Minus one!\n");
        break;
    default:
        printf("Not minus one!\n");
}

Why is it that the program prints Minus One! ? Is num casted to a size_t in the switch statement? Is this behavior defined?

Answer 1

From the C standard on switch:

6.8.4.2 The switch statement
...
Semantics
...
5 The integer promotions are performed on the controlling expression. The constant expression in each case label is converted to the promoted type of the controlling expression. If a converted value matches that of the promoted controlling expression, control jumps to the statement following the matched case label.

size_t is also an unsigned type per the standard (6.5.3.4 The sizeof operator, Semantics, 4).

So, your -1 is converted to size_t in both size_t num = -1; and case -1: . No wonder that (size_t)-1==(size_t)-1 evaluates to true.

Answer 2

Yes, this behavior is defined. The values are cast to size_t as you'd expect. From C99 §6.8.4.2/5:

The integer promotions are performed on the controlling expression. The constant expression in each case label is converted to the promoted type of the controlling expression. If a converted value matches that of the promoted controlling expression, control jumps to the statement following the matched case label.

The "controlling expression" is the expression appearing inside the parentheses after the switch keyword, ie num in this case.

So, the -1 in the case label is converted to size_t (the promoted type of num ). Since the value of num is just that, the code in that case statement is executed.