Thank you for being one of the few people who understands that in C/C++, `unsigned OP unsigned` can have each operand be promoted to a signed integer and then have the operation overflow and cause undefined behavior.

I chose to deal with this problem by doing a "pointless" operation to force a promotion to at least unsigned int. For example:

    uint16_t x = 0xFFFF;
    uint16_t y = 0xFFFF;
    uint16_t z = (uint16_t)((x + 0U) * (y + 0U));

This piece of code will work on any machine, such as: (uint16_t = unsigned short = 16 bits, uint32_t = unsigned int = 32 bits); (uint16_t = unsigned short = unsigned int = 16 bits, uint32_t = unsigned long = 32 bits).

> stuff like hashcode computations using uint32_t with multiplications, relying on the C standard guaranteeing wraparound for unsigned overflows. But with 64-bit int, uint32_t will promote to a signed int, and overflows will thus be undefined behavior.

Yeah, except that multiplying two 32-bit values, recast as 64-bit signed integers, will not overflow. Even adding another 32-bit value to this product will not overflow. Throw in the final cast to uint32_t to throw away the upper sign bits, and you get the identical result.