How to build n bit with 1?

[CakeByTheOcean]

I want a function, which takes a int (not a compile time constant).
Gives me back a uint64_t, which has so many 1 as the int parameter.

Is that possible and fast?

  f(1); // 1
  f(5); // 11111
  f(10); // 1111111111  

[mbozzi]

#include <cstdlib>
#include <iostream>

constexpr std::uint64_t repeat_one(int n)
{
  constexpr std::uint64_t lut[] = 
  {
    UINT64_C(0),
    UINT64_C(1),                   UINT64_C(11),
    UINT64_C(111),                 UINT64_C(1111),
    UINT64_C(11111),               UINT64_C(111111), 
    UINT64_C(1111111),             UINT64_C(11111111),
    UINT64_C(111111111),           UINT64_C(1111111111),
    UINT64_C(11111111111),         UINT64_C(111111111111),
    UINT64_C(1111111111111),       UINT64_C(11111111111111),
    UINT64_C(111111111111111),     UINT64_C(1111111111111111),
    UINT64_C(11111111111111111),   UINT64_C(111111111111111111),
    UINT64_C(1111111111111111111), UINT64_C(11111111111111111111)
  };
  
  return lut[n];
}

int main()
{
    for (int i = 0; i <= 20; ++i)
        std::cout << repeat_one(i) << '\n';   
}

[dutch]

This does 1 to 64. I can't think of a better way of handling zero than just adding a specific test for it. But if you don't need 0 and you don't need error checking (for negative values or those over 64) then:

uint64_t f(int n) {
    return uint64_t(-1) >> (64 - n);
}

[Duthomhas]

Er, basic bit manipulation:

    (1U << n) - 1

constexpr uint64_t repeat_one( int n )
{
  return ((uint64_t)1 << n) - 1;
}

Remember, unsigned rollover IS defined, so this is guaranteed work even if you ask for 64 ones.

[dutch]

Er, not quite.
Remember that shifting by the number of bits or greater is undefined.
Asking for 64 yields 0 for me.

int main() {
    auto x = 1ULL << 64;  // gives a warning
}

[Duthomhas]

Oops! I forgot that.

constexpr uint64_t repeat_one( int n )
{
  return ((uint64_t)1 << (n > 64 ? n : 0)) - 1;
}