If n != 0 and -1 cast to unsigned size_t divided by n is less than m. So that m times n would be bigger than size_t’s maximum value.

Edit: inverted meaning since check is for error, not for ok case.

You want to evaluate p in:

  p = m n

If that is gonna overflow, p > SIZE_MAX.

Then it would be that (by substitution)

  m n > SIZE_MAX

so

  m > SIZE_MAX/n

except if n = 0. In the latter case, m can be as big as it wants--p will always evaluate to 0.

And (size_t)-1 is a silly way to write SIZE_MAX (but it saves you a dependency).

Thanks! Indeed the `(size_t)-1` threw me off, and the `m && n` was an interesting obfuscation for checking whether either are 0.