Narrowing conversion from char to uint8_t
c++, c++11, narrowing
Solution
Although char doesn't necessarily have to be 8-bits long, that's not the problem here. You are converting from `signed char` to unsigned (`uint8_t`), that's the reason for the error.
This:
const int8_t foo[] = {
'\xf2'
};
will compile fine.
Problem
Compiling the following snippet using C++11(demo here): ``` #include <stdint.h> int main() { const uint8_t foo[] = { '\xf2' }; } ``` Will trigger a warning(at least on GCC 4.7), indicating that there's a narrowing conversion when converting `'\xf2'` to `uint8_t`. Why is this? `sizeof(char)` is always `1`, which should be the same as `sizeof(uint8_t)`, shouldn't it? Note that when using other char literals such as `'\x02'`, there's no warning.