INT_MAX and INT16_MAX are not interchangeable. > I wonder if the compiler just used the full register. I'll have to check the disassembly. No need, you'd figu

INT_MAX and INT16_MAX are not interchangeable. > I wonder if the compiler just used the full register. I'll have to check the disassembly. No need, you'd figure this out just by knowing what the C spec says and by knowing what the value of INT_MAX is. When you compute the remainder in C, the operands first undergo "integer promotion" and then go through the "usual arithmetic conversions". The result is that you end up with two values of the same type, no narrower than int. So in C, the following function: short broken_modulo(short x, short y) { return ((x % y) + y) % y; } Is equivalent to the following, with the implicit casts made explicit: short broken_modulo(short x, short y) { int ix = (int)x; int iy = (int)y; // All int, no short. int result = ((ix % iy) + iy) % iy; return (short)result; } Note that this conversion happens in ALL C implementations, because it is mandated by the spec, and no implementation-defined stuff is happening here. Do note that it is possible for int to be 16-bit, e.g., // Possible. typedef int int16_t; It's just uncommon to see that in this millennium. I used "int" and "short" above rather than "int16_t" because the exact rank of "int16_t" vs "int" with regard to arithmetic conversions is implementation dependent, but the rank of "short" vs "int" is not implementation dependent.