How is pow() calculated in C?

Question

Our professor said that you can\'t calculate a^b if a<0 using pow() because pow() uses natural logarithms to calculate it (a^b

Answer 1

Assuming an x86 series processor, pow is the equivalent of

double pow(double base, double exp)
{
   return exp2(exp * log2(base));
}

Where exp2 and log2 are CPU primitives for the exponential and logarithm operations in base 2.

Different CPUs inherently have different implementations.

In theory if you didn't have pow you could write:

double pow(double base, double exponent)
{
   return exp(exponent * log(base));
}

but this loses precision over the native version due to accumulative roundoff.

And Dietrich Epp revealed I missed a bunch of special cases. Nevertheless I have something to say about roundoff that should be allowed to stand.