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

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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 );

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