CODEWITHSUNDEEP
cprintf
DevSolar
DevSolar

Reputation: 70223

printf(), unsigned char, and %hhd

The program:

#include <stdio.h>
#include <limits.h>

int main( void )
{
#ifdef __CHAR_UNSIGNED__
    printf( "%d\n", __CHAR_UNSIGNED__ );
#endif
    printf( "%d\n", CHAR_MAX );
    printf( "%d\n", CHAR_MIN );
    printf( "%hhd\n", CHAR_MAX );
}

Output (on my x86_64 desktop):

127
-128
127

That is as expected. Now, I ran the same on a Raspberry Pi (ARM):

1
255
0
-1

So... apparently I have misunderstood some step CHAR_MAX is taking in its way to output, as the output I was expecting in that last line -- passing CHAR_MAX to %hhd on a machine with char being unsigned -- would have been 255.

(And if you should ask, the result is the same for (char)CHAR_MAX, and (unsigned char)CHAR_MAX.)

What am I missing?

This happened to me during regression testing my own printf() implementation on the Raspberry Pi -- which, by the way, does print 255. So... only one implementation is doing it right, and I have the sinking feeling it's not mine...

Upvotes: 1

Views: 7084

Answers (1)

Some programmer dude
Some programmer dude

Reputation: 409136

The format "%hhd" is for a signed byte. If you pass an unsigned byte then you technically have undefined behavior. To print an unsigned byte you need to use "%hhu".

What really happens is because of how two's complement integers work, and that 0xff is equal to -1.

Upvotes: 8

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