Printing uint8_t variables using std::cout in C++ [duplicate]

问题

This question already has answers here:

uint8_t can't be printed with cout (8 answers)

Closed 5 years ago.

In the following code, it appears that std::cout does not print the variable of type uint8_t properly.

int main() {
    uint8_t var = 16;
    std::cout << "value: " << var << std::endl;
}

output:

value: 

I don't have problem with similar uintX_t types.

I know I'm missing something here, but I can't figure what it is.

回答1:

uint8_t maps to unsigned char on most systems. As the result, the override that interprets 16 as a non-printable character gets invoked. Add a cast to int to see the numeric value:

std::cout << "value: " << (int)var << std::endl;

来源：https://stackoverflow.com/questions/30856861/printing-uint8-t-variables-using-stdcout-in-c