Printing hexadecimal characters in C

You are seeing the ffffff because char is signed on your system. In C, vararg functions such as printf will promote all integers smaller than int to int. Since char is an integer (8-bit signed integer in your case), your chars are being promoted to int via sign-extension.

Since c0 and 80 have a leading 1-bit (and are negative as an 8-bit integer), they are being sign-extended while the others in your sample don't.

char    intc0 -> ffffffc080 -> ffffff8061 -> 00000061

Here's a solution:

char ch = 0xC0;printf("%x", ch & 0xff);

This will mask out the upper bits and keep only the lower 8 bits that you want.

Indeed, there is type conversion to int.Also you can force type to char by using %hhx specifier.

printf("%hhX", a);

In most cases you will want to set the minimum length as well to fill the second character with zeroes:

printf("%02hhX", a);

ISO/IEC 9899:201x says:

7 The length modifiers and their meanings are: hh Specifies that a following d, i, o, u, x, or X conversion specifier applies to a signed char or unsigned char argument (the argument will have been promoted according to the integer promotions, but its value shall be converted to signed char or unsigned char before printing); or that a following

You can create an unsigned char:

unsigned char c = 0xc5;

Printing it will give C5 and not ffffffc5.

Only the chars bigger than 127 are printed with the ffffff because they are negative (char is signed).

Or you can cast the char while printing:

char c = 0xc5; printf("%x", (unsigned char)c);