Narrowing conversion from double to float: is overflow behaviour guaranteed?

Question

If I try this

float f = (float)numeric_limits<double>::infinity();

Or indeed, try to cast anything bigger than float max down to a float, am I guaranteed to end up with infinity?

It works on GCC, but is it a standard though?

Answer 1

float f = (float)numeric_limits<double>::infinity();

This is guaranteed to set f to infinity if your compilation platform offers IEEE 754 arithmetic for floating-point computations (it usually does).

Or indeed, try to cast anything bigger than float max down to a float, am I guaranteed to end up with infinity?

No. In the default IEEE 754 round-to-nearest mode, a few double values above the maximum finite float (that is, FLT_MAX) convert to FLT_MAX. The exact limit is the number midway between FLT_MAX (0x1.fffffep127 in C99 hexadecimal representation) and the next float number that could be represented if the exponent in the single-precision format had a larger range, 0x2.0p127. The limit is thus 0x1.ffffffp127 or approximately 3.4028235677973366e+38 in decimal.