Is the inverse of std::numeric_limits::infinity() zero?

Question

Is there anything in the C++ standard (or the IEEE 754 floating-point standard) that guarantees that 1./std::numeric_limits<double>::infinity() is zero (or at least a small number)?

Answer 1

Any finite number divided by infinity results in zero under IEEE 754 (and therefore the same in most typical C++ implementations).

If the sign of the of numerator and denominator differ, the result will be negative zero, which is equal to zero.