C++ – Difference Between int64 and int64_t in C++

From the friendly Wikipedia:

The stdlib.h and stddef.h header files define a datatype called size_t which is used to represent the size of an object. Library functions that take sizes expect them to be of type size_t, and the sizeof operator evaluates to size_t.

The actual type of size_t is platform-dependent; a common mistake is to assume size_t is the same as unsigned int, which can lead to programming errors, particularly as 64-bit architectures become more prevalent.

Also, check Why size_t matters

You don't need to go to 64-bit to see something like this. Consider int32_t on common 32-bit platforms. It might be typedef'ed as int or as a long, but obviously only one of the two at a time. int and long are of course distinct types.

It's not hard to see that there is no workaround which makes int == int32_t == long on 32-bit systems. For the same reason, there's no way to make long == int64_t == long long on 64-bit systems.

If you could, the possible consequences would be rather painful for code that overloaded foo(int), foo(long) and foo(long long) - suddenly they'd have two definitions for the same overload?!

The correct solution is that your template code usually should not be relying on a precise type, but on the properties of that type. The whole same_type logic could still be OK for specific cases:

long foo(long x);
std::tr1::disable_if(same_type(int64_t, long), int64_t)::type foo(int64_t);

I.e., the overload foo(int64_t) is not defined when it's exactly the same as foo(long).

[edit] With C++11, we now have a standard way to write this:

long foo(long x);
std::enable_if<!std::is_same<int64_t, long>::value, int64_t>::type foo(int64_t);

[edit] Or C++20

long foo(long x);
int64_t foo(int64_t) requires (!std::is_same_v<int64_t, long>);