Is casting of infinity to integer undefined?

Question

Is the casting of infinity (represented by float) to an integer an undefined behavior?

The standard says:

4.10 Floating-integral conversions

A prvalue of a floating point type can be converted to a prvalue of an integer type. The conversion truncates; that is, the fractional part is discarded. The behavior is undefined if the truncated value cannot be represented in the destination type.

but I can't tell whether "truncated value cannot be represented" covers infinity.

I'm trying to understand why std::numeric_limits<int>::infinity() and static_cast<int>(std::numeric_limits<float>::infinity() ) have different results.

#include <iostream> #include <limits>  int main () {     std::cout << std::numeric_limits<int>::infinity () << std::endl;     std::cout << static_cast<int> (std::numeric_limits<float>::infinity () ) << std::endl;     return 0; } 

Output:

0   -2147483648   

The result of std::numeric_limits<int>::infinity() is well defined and equal to 0, but I can't find any information about casting infinity.

Answer 1

Casting of infinity to integer is undefined.

The behavior is undefined if the truncated value cannot be represented in the destination type.

Says it all. Since truncation removes precision but not magnitude, a truncated infinity is still infinity and integers cannot represent infinity.