Draining the entropy of std::random_device

Question

I'm using std::random_device and would like to check for its remaining entropy. According to cppreference.com:

std::random_device::entropy

double entropy() const noexcept;

[...]

Return value

The value of the device entropy, or zero if not applicable.

Notes

This function is not fully implemented in some standard libraries. For example, LLVM libc++ always returns zero even though the device is non-deterministic. In comparison, Microsoft Visual C++ implementation always returns 32, and boost.random returns 10.

The entropy of the Linux kernel device /dev/urandom may be obtained using ioctl RNDGETENTCNT - that's what std::random_device::entropy() in GNU libstdc++ uses as of version 8.1

So under Linux ang g++ >= 8.1, I should be good... but I'm not:

#include <random>
#include <iostream>

void drain_entropy(std::random_device& rd, std::size_t count = 1)
{
    while (count --> 0) {
        volatile const int discard = rd();
        (void) discard;
    }
}

int main()
{
    std::random_device rd;
    std::cout << "Entropy: " << rd.entropy() << '\n'; // Entropy: 32
    drain_entropy(rd, 1'000'000);
    std::cout << "Entropy: " << rd.entropy() << '\n'; // Entropy: 32
}

Live demo on Coliru (which runs under Linux, right?)

I'm expecting that generating numbers from the device drains its entropy. But it doesn't.

What's happening?

Answer 1

The library will not return an entropy value greater than the number of bits in its result type, which is 32 in this case.

See libstd code:

const int max = sizeof(result_type) * __CHAR_BIT__;
if (ent > max)
  ent = max;

The documentation you linked to explains this:

Obtains an estimate of the random number device entropy, which is a floating-point value between 0 and log 2(max()+1) (which is equal to std::numeric_limits::digits).