Casting from std::size_t to void

Question

In a lot of older code, I've seen variations on the following snippet:

std::size_t  some_size = some_function_that_returns_size_t();
(void)some_size;
assert(some_size > some_other_size);

What is the purpose of the cast to void ?

Bear in mind this is not casting to void*

If I has to guess, this looks like a safety check to ensure that the machine size_t is larger than a void at compile time.

Playing around in an interpreter reveals that the cast will return a value different than some_size if the initial value is large enough, but since it's not being assigned, I honestly don't see the point.

In case it matters, this was found in an allocation library as part of the allocation checks against alignment and size.

Answer 1

据我所知，这种技巧是用来防止编译器发出警告，指出已定义但未使用变量。