在 C++ 中以十六进制格式打印 unsigned char(BYTE)。 使用 std::cout

Question

I've looked through all previous version of this topic and none have seemed to help. I've got a value located in unsigned char called Offset. When printing this with printf("%02X\\n", Offset); It prints out the correct value which is 0x2B. Though when I try with std::cout << std::hex << Offset << std::endl; It always prints a different value. I have also tried:

std::cout << std::hex << (unsigned int)Offset << std::endl;
std::cout << std::hex << static_cast<unsigned>(Offset) << std::endl;

Though none have seemed to print out the correct value.

Answer 1

As you defined "Offset" like this unsigned char Offset = 0x2B; , Offset's type now is unsigned char and you cant not print a unsigned char variable as a std::hex . You must first convert it to the std::hex type then print it.

#include <iostream >


int main() {

    unsigned char Offset = 0x2B;
    printf("%02X\n", Offset);
    std::cout << std::hex << Offset << std::endl;
    std::cout << std::hex << (unsigned int)Offset << std::endl;
    std::cout << std::hex << static_cast<unsigned>(Offset) << std::endl;
    return 0;
}