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 🙂 .


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.

Leave a Comment

Hata!: SQLSTATE[HY000] [1045] Access denied for user 'divattrend_liink'@'localhost' (using password: YES)