According to http://en.cppreference.com/w/cpp/numeric/math/pow , when std::pow is used with integer parameters, the result is promoted to a double.
My question is then the following:
How safe is to compare an integer type with the result of a std::pow(int1, int2)?
For example, can the if below evaluate to true?
std::size_t n = 1024;
if(n != std::pow(2, 10))
cout << "Roundoff issues..." << endl;
That is, is it possible that the result on the rhs can be something like 1023.99...9 so when converted to size_t becomes 1023?
My guess is that the response in a big NO, but would like to know for sure. I am using these kind of comparisons when checking for dimensions of matrices etc, and I wouldn't like to use a std::round everywhere.
It is funny you should ask, because someone else on StackOverflow had a question that was caused by the very fact that pow applied to small integers did not compute the obvious result on their platform (see also my writeup).
So yes, when applying pow to small integers, both arguments and ideal mathematical result are exactly representable. This does not force the implementation of exp to return the mathematical result, because no standard specifies that pow cannot be inaccurate by more than one ULP. And at least one very popular platform provides by default a pow function that does not compute pow(10, 2) as 100, but you are free to take you chances with pow(2, N) and perhaps it will happen to always return the integer you are entitled to expect.