uint8_t can't be printed with cout

Question

I have a weird problem about working with integers in C++.

I wrote a simple program that sets a value to a variable and then prints it, but it is not working as expe

Answer 1

As others said before the problem occurs because standard stream treats signed char and unsigned char as single characters and not as numbers.

Here is my solution with minimal code changes:

uint8_t aa = 5;

cout << "value is " << aa + 0 << endl;

Adding "+0" is safe with any number including floating point.

For integer types it will change type of result to int if sizeof(aa) < sizeof(int). And it will not change type if sizeof(aa) >= sizeof(int).

This solution is also good for preparing int8_t to be printed to stream while some other solutions are not so good:

int8_t aa = -120;

cout << "value is " << aa + 0 << endl;
cout << "bad value is " << unsigned(aa) << endl;

Output:

value is -120
bad value is 4294967176

P.S. Solution with ADL given by pepper_chico and πάντα ῥεῖ is really beautiful.