Initialize integer literal to std::size_t

There is no such standard facility. C99 and C++11 implementations do have such macros in <stdint.h>/<cstdint>. But even there, the macros are only defined for the stdint.h types, which do not include size_t.

You could define a user-defined literal operator:

constexpr std::size_t operator "" _z ( unsigned long long n )
    { return n; }

auto sz = 5_z;
static_assert( std::is_same< decltype( sz ), std::size_t >::value, "" );

The constexpr is necessary to use it in array bounds int arr[ 23_z ] or case 9_z: labels.

Most would probably consider the lack of macros to be an advantage :) .

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Cuteness aside, the best way is to use brace initialization: std::size_t{ 42 }. This is not equivalent to std::size_t( 42 ) which is like a nasty C cast — presumably what you were avoiding with static_cast. Quite the opposite: the braces require that the value inside is exactly representable in the targeted type. So, char{ 300 } and std::size_t{ -1 } are both ill-formed.

Braces and parens look similar, but they're polar opposites in safety when initializing temporaries. Braces are safer than the literal operator could ever be, since unlike a function they can discriminate compile-time values.

There is no dedicated suffix for std::size_t. In C++11, you could create a user-defined literal for it, though:

std::size_t operator "" _sz (unsigned long long int x)
{
  return x;
}

// Usage:

auto s = 1024_sz;

static_assert(std::is_same<decltype(s), std::size_t>::value, "He's wrong");

Live example

Depending on the function, you may also be able to do this and may find it cleaner:

auto result = func<size_t>(1, some_var);

For example, I've done this with std::max:

auto result = std::max<size_t>(0, std::min<size_t>(index, vec.size()-1));

By explicitly specifying the template instantiation, the casts can be implicit. However, do note that this is a cast, so is susceptible to errors which Potatoswatter's brace initialization isn't.