printf adds extra `FFFFFF` to hex print from a char array [duplicate]

Solution 1:

Sign extension. Your compiler is implementing char as a signed char. When you pass the chars to printf they are all being sign extended during their promotion to ints. When the first bit is a 0 this doesn't matter, because it gets extended with 0s.

0xAF in binary is 10101111 Since the first bit is a 1, when passing it to printf it is extended with all 1s in the conversion to int making it 11111111111111111111111110101111, the hex value you have.

Solution: Use unsigned char (instead of char) to prevent the sign extension from occurring in the call

const unsigned char raw[] = {0x20,0x00,0xAF,0x00,0x69,0x00,0x33,0x00,0x5A,0x00};

All of these values in your original example are being sign extended, it's just that 0xAF is the only one with a 1 in the first bit.

Another simpler example of the same behavior (live link):

signed char c = 0xAF; // probably gives an overflow warning
int i = c; // extra 24 bits are all 1
assert( i == 0xFFFFFFAF );

Solution 2:

That's because 0xAF when converted from a signed character to a signed integer is negative (it is sign extended), and the %02X format is for unsigned arguments and prints the converted value as FFFFFFAF.

The extra characters appear because printf %x will never silently truncate digits off of a value. Values which are non-negative get sign extended as well, but that's just adding zero bits and the value fits in 2 hex digits, so printf %02 can do with a two digit output.

Note that there are 2 C dialects: one where plain char is signed, and one where it is unsigned. In yours it is signed. You may change it using an option, e.g. gcc and clang support -funsigned-char and -fsigned-char.