sprintf unexpected result for int8_t

Asked Mar 03 '17 at 13:03

Active Mar 03 '17 at 13:03

Viewed 131 times

I tried to write hexadecimal value of int8_t variable to a buffer with the following code:

#include <stdint.h>
#include <stdio.h>

int main(){

    char buf[16];
    uint16_t val = 0xDEAD;
    int8_t val2 = -50;

    sprintf(buf, "%04x%02x", val, val2);
    printf("%s\n", buf);

    sprintf(buf, "%04x%02x", val, (uint8_t)val2);
    printf("%s\n", buf);

    return 0;

}

And the output is as follows:

deadffffffce
deadce

How is the first result generated?

I compiled with: gcc (Ubuntu 6.2.0-5ubuntu12) 6.2.0 20161005

asked Mar 03 '17 at 13:03

pbn

2,406
2
26
39

Because that format specification does not *truncate* if the result is larger than the requested field width. `val2` is signed, and negative, so it is sign-extended when it is promoted to `int` and passed to `sprintf`. – Weather Vane Mar 03 '17 at 13:07
Try to use the "%hhx" specification for the unit8_t value. See also [here](http://stackoverflow.com/questions/4586962/what-is-the-purpose-of-the-h-and-hh-modifiers-for-printf). – Willy K. Mar 03 '17 at 13:23
@WillyK.: That's not the correct way! – too honest for this site Mar 03 '17 at 13:24
1

Your code invokes undefined behaviour. To print `stdint.h` types, use the macros from `inttypes.h`. There is no guarantee they fixed-width types use the types you assume. Worse: you pass a signed type, but `%x` expects an unsigned integer. – too honest for this site Mar 03 '17 at 13:25
It is shorter, but equally valid to cast the `int8_t` to `unsigned`... – Antti Haapala -- Слава Україні Mar 03 '17 at 13:33
*that is to say: to `uint8_t`. – Antti Haapala -- Слава Україні Mar 03 '17 at 13:41
"How is the first result generated?" --> curious, what did you expect the first result to be? – chux - Reinstate Monica Mar 03 '17 at 16:17

sprintf unexpected result for int8_t

0 Answers0