Why is std::size_t 4 bytes on 32bit systems when unsigned long long is 8 bytes on both 32bit and 64 bit systems?

Question

Question is quite simple.

On 32bit systems:

std::cout << sizeof(unsigned int);        //4
std::cout << sizeof(unsigned long long);  //8
std::cout << sizeof(std::size_t);         //4

On 64bit systems:

std::cout << sizeof(unsigned int);        //4
std::cout << sizeof(unsigned long long);  //8
std::cout << sizeof(std::size_t);         //8

I only checked the implementation that MSVC has, and it looks like this:

#ifdef _WIN64
    typedef unsigned __int64 size_t;
#else
    typedef unsigned int     size_t;
#endif

So why not make std::size_t unsigned long long (std::uintmax_t) on both 32bit and 64bit systems when they clearly support it? Or am I wrong in that?

Answer 1

The point of size_t is to be able to hold the size of the biggest possible object. On a 32 bit system no object can occupy more than 2**32 bytes, so a 32 bit type is sufficient.

To use a 64 bit type would be wasteful of space and potentially more expensive in run time.

Answer 2

That would be a pointless waste. On a 32 bit machine you have a 4 GB address space, so you cannot have objects bigger than 4 GB, so the range of a 32 bit size_t is perfectly adequate.