sizeof(x++) does not increase x

I thought x would increase to 6 if I run the code below, so 4 and 6 would come out When I printed it out as dev c++ in the window, 4 and 5 came out.

Why is the x value still 5 when sizeof(x++) increased?

#include <stdio.h>

int main() {
    int x = 5;
    printf("%d and ", sizeof(x++)));
    printf("%d\n", x);
    return 0;
}

Output: 4 and 5

c sizeof

2022-09-21 20:04

1 Answers

In 6.5.3.4/2 of C99 Standard,

The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.

That's what it says.

sizeof() may be an operand with the name of exression or type (int, char, etc.) The result returns the size of the operand type

At this time, calculate the operand only when it is of variable length array type. In other words, x is int type and does not correspond to it, so x++ does not run.

2022-09-21 20:04

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