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c++undefined-behavior

Is casting of infinity to integer undefined?

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.

Solution

  • You said

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

    but it all boils down to

    What is the result of truncating infinity.

    The C standard (incorporated into C++ via 26.9) answers that quite plainly:

    C standard semantics for <code>trunc</code>

    Since truncation of infinity is still infinity, and infinity cannot be represented in int (I hope there's no question about this part), the behavior is undefined.