uint32_t destructor return value

Question

Today I saw this code, inside a class:

static const uint32_t invalid_index = ~uint32_t();

My question is, what is the return value of a uint32_t destructor, and why is it useful?

Answer 1

That's not a destructor, but a bitwise NOT operator applied to a value-initialized uint32_t .

A value initialized integral type is 0 , so you're taking the bitwise NOT of 0 .

Similar to:

uint32_t x = uint32_t();  // 32 0's in binary form
uint32_t y = ~x;          // 32 1's in binary form

Answer 2

First of all, as many have already mentioned, the code you saw,

static const uint32_t invalid_index = ~uint32_t();

is not a destructor call but the bitwise "not" ~ , applied to the default value of the type, uint32_t() , ie ~(uint32_t(0)) .

Now to your question,

My question is, what is the return value of a uint32_t destructor, and why is it useful?

The return type of the pseudo-destructor (it's not a real destructor, just a do-nothing operation with the same notation as a destructor call) is void , and it's mainly useful for generic programming where you don't know the type.

Example:

uint32_t x;
x.~uint32_t();  // Silly but valid, a pseudo-destructor call.

Answer 3

It is not a destructor, it is binary not. Here the invalid index is equal to ~uint32_t(0). Which is a 32 bit unsigned integer with all bits set. ie, 0xffffffff.

Answer 4

它是按位NOT ，它可用于找到1的补码（例如~1011 = 0100）或作为尝试找到2s补码的中间步骤（例如[~1011] + 0001 = 0101）。