I want to round a double to the nearest integer and round to even in halfway cases. Per documentation std::nearbyintf should do this:

If the current rounding mode is FE_TONEAREST, this function rounds to even in halfway cases (like std::rint, but unlike std::round).

Another function that does this is std::rintf. The docs of std::rintf state:

The only difference between std::nearbyint and std::rint is that std::nearbyint never raises FE_INEXACT.

So, I did expect the same results for both functions, but they return different results for double x = 3.4999999.

The following reprex demonstrates this:

#include <iostream>
#include <iomanip>
#include <cfenv>
#include <cmath>

int main() {
  std::fesetround(FE_TONEAREST);

  std::cout << std::setprecision(15)
            << " std::nearbyintf(3.5) = " << std::nearbyintf(3.5) << "n"
            << " std::rint(3.5) = " << std::rint(3.5) << std::endl;

  std::cout << std::setprecision(15)
            << " std::nearbyintf(2.5) = " << std::nearbyintf(2.5) << "n"
            << " std::rint(2.5) = " << std::rint(2.5) << std::endl;

  std::cout << std::setprecision(15)
            << " std::nearbyintf(3.4999999) = " << std::nearbyintf(3.4999999) << "n"
            << " std::rintf(3.4999999) = " << std::rint(3.4999999) << std::endl;
            
  return 0;
}

This example can be run here and produces the following output:

 std::nearbyintf(3.5) = 4
 std::rint(3.5) = 4
 std::nearbyintf(2.5) = 2
 std::rint(2.5) = 2
 std::nearbyintf(3.4999999) = 4
 std::rintf(3.4999999) = 3

Any advice on why they differ would be greatly appreciated.