C-将char显示为十六进制

Question

I want to print the ASCII code of a char in hex; for example, for

char a = 0xA5;

I want to print A5 on the console. Here is what I have tried:

char a = 0xA5;
printf("%02X", a);

but i get FFFFFFA5 . How could I solve this?

Answer 1

Cast the value to unsigned char , then cast again to unsigned int to be printed via %X .

char a = 0xA5;
printf("%02X", (unsigned int)(unsigned char)a);

Note that conversion to signed integer which is not capable to store original value is implementation-defined, but conversion to unsigned integer is defined, according to N1256 6.3.1.3

Answer 2

The "problem" is called sign extension -- parameteres are passed as int by default so a char would be converted to int and in the process the sign extension would means that the extra f are added -- make the char unsigned like this

unsigned char a = 0xA5;
printf("%02X", a);

and the compiler will understand how to treat your data.