What is the return type of sizeof operator?

Question

What is the return type of sizeof operator? cppreference.com & msdn says sizeof returns size_t. Does it really return a size_t? I'm using VS2010 Professional, and targeting for x64.

int main() {     int size   = sizeof(int);     // No warning     int length = strlen("Expo");  //warning C4267: 'initializing' : conversion from 'size_t' to 'int', possible loss of data     return 0; } 

I have this question because first line is not issuing any warning, whereas the second does. Even if I change it to char size, I don't get any warnings.

Answer 1

C++11, §5.3.3 ¶6

The result of sizeof and sizeof... is a constant of type std::size_t. [ Note: std::size_t is defined in the standard header (18.2). — end note ]

You can also do a quick check:

#include <iostream> #include <typeinfo> #include <cstdlib>  int main() {     std::cout<<(typeid(sizeof(int))==typeid(std::size_t))<<std::endl;     return 0; } 

which correctly outputs 1 on my machine.

As @Adam D. Ruppe said in the comment, probably the compiler does not complain because, since it already knows the result, it knows that such "conversion" is not dangerous

Answer 2

size_t is an alias of some implementation-defined unsigned integral type. In C++ opposite to C where sizeof operator may be applied to VLA arrays the operand of sizeof operator is not evaluated (at run time). It is a constant. If the value of sizeof operator can be fit into int type the compiler does not issue a warning. In the second example std::strlen is evaluated at run time so its result can do not fit into int so the compiler issues a warning. You could substitute std:;strlen with your own constexpr function (some recursive function). In this case if the result can fit into int I think that the compiler will not issue a warning.