Under http://www.cpp.sh long same size as long long

[MikeStgt]

// Test long vs long long
#include <iostream>

int main()
{
    unsigned long long x(~0), y(0), z(x);
    for (; z > 0; ++y, z /= 2);
    std::cout << x << " = 2^" << y << "-1";
}

long long or long only makes not difference. Or is there an option I've overlooked?

[jonnin]

some systems, they are the same.

[Peter87]

The size depends on your system/compiler.

int is at least 16 bits.
long is at least 32 bits.
long long is at least 64 bits.

[MikeStgt]

long long is at least 64 bits.

With long it's already 64 bits , so I hoped to get with long long a bit more, or two ;)

[dutch]

On gcc you can include <cstdint> and try __int128_t or __uint128_t.
I don't think cout knows how to output them, though, so you may have to write your own output routine.

[Ganado]

For arbitrarily greater than 64-bit integers, one must look into "BigInteger" or "Multiprecision" libraries.

Alternatively, if only up to 128-bit integers are needed, gcc and clang have __int128 (if supported on your system), and boost has its version of int128 as well.
https://stackoverflow.com/a/35026162/8690169 (gcc/clang __int128)
https://stackoverflow.com/a/18439628/8690169 (boost int128)

[MikeStgt]

For arbitrarily greater than 64-bit integers, one must look into "BigInteger" or "Multiprecision" libraries.

I just try to fathom the "plain" system without additional libraries. It's for fun, not for advantage -- silly as speed driving.

[dutch]

silly as speed driving

Are you just a random phrase generator?

[MikeStgt]

On gcc you can include <cstdint> and try __int128_t or __uint128_t.

// Test __uint128_t vs long long
#include <iostream>
#include <cstdint>

int main()
{
    __uint128_t x(~0), z(x);
    unsigned y(0);
    for (; z > 0; ++y, z /= 2);
    std::cout << "x = 2^" << y << "-1";
}

Works, tnx.