从值小于256的uint16到uint8的字节序安全转换

Question

I am interacting with an API that returns uint16_t values; in this case I know that the value is never going to exceed 255. I need to convert the value to a uint8_t for usage with a separate API. I am currently doing this in the following way:

uint16_t u16_value = 100;
uint8_t u8_value = u16_value << 8;

This solution currently exposes endianness issues if moving from a little-endian (my current system) to a big-endian system.

What is the best way to mitigate against this?

Answer 1

From cppreference

For unsigned and positive a, the value of a << b is the value of a * 2**b , reduced modulo maximum value of the return type plus 1 (that is, bitwise left shift is performed and the bits that get shifted out of the destination type are discarded).

There's nothing about endianness here. You can just do

uint16_t u16_value = 100;
uint8_t u8_value = u16_value;

or

uint16_t u16_value = 100;
uint8_t u8_value = static_cast<uint8_t>(u16_value);

To be explicit.