Seemingly compiler-specific failure to convert 64 bit bitset to binary string representation

Question

Here is my code:

#include<iostream>
#include<string>
#include<bitset>

using namespace std;

int main()
{
    long long int number;
    number = 123456789123456789;
    string binary = bitset<64>(number).to_string();
    cout<<binary<<"\n";
    return 0;
}

Here is the result: 0000000000000000000000000000000010101100110100000101111100010101 but it's wrong.

Info from comments (and OPs experiment):
Same code results in expected binary representation in other environments.

What is the reason?

Answer 1

I can reproduce the problem under the following circumstances:

• 32-bit target architecture,
• either old GCC version or compilation against C++03.

For example, with GCC 10.2, this happens with -m32 -std=c++03, with GCC 4.9.2 just with -m32. Until C++11, there was no unsigned long long defined by the standard. Though C++98/03 implementations may provide it as a non-standard extension, the parameter of the constructor of std::bitset was of unsigned long type only. Which, in the above-described cases, is only 32-bit long. This is where you lost the upper bits.

Live demo: https://godbolt.org/z/PWdY8K

The relevant part in libstdc++ is here:

#if __cplusplus >= 201103L
  constexpr bitset(unsigned long long __val) noexcept
  : _Base(_Sanitize_val<_Nb>::_S_do_sanitize_val(__val)) { }
#else
  bitset(unsigned long __val)
  : _Base(__val)
  { _M_do_sanitize(); }
#endif