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?
From cppreference
For unsigned and positive a, the value of
a << b
is the value ofa * 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.
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