How to resolve GCC compilation [-Wformat] warning?

Question

I am using this approach to convert hex string to byte array.The code is works correct. While compiling this code I am getting below compilation warning. Is there any way I can resolve it?

/* test.c */
#include <stdio.h>
#include <stdint.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
    if (argc < 2)
    {
        printf("Usage: ./test <input hex string>\n");
        return 0;
    }

    char *hexstring = argv[1];
    printf("hextring:%s\n", hexstring);
    uint8_t str_len = strlen(hexstring);
    printf("length:%d\n", str_len);
    uint8_t array_size = str_len / 2;
    printf("array_size:%d\n", array_size);
    uint8_t *input = (uint8_t *)calloc(array_size, sizeof(uint8_t));

   for (int i = 0; i < array_size; i++)
   {
       sscanf(hexstring, "%02x", &input[i]);
       hexstring = hexstring + 2;
   }
   for (int i = 0; i < array_size; i++)
   {
       printf("input[%d]:[%.2x]\n", i, input[i]);
   }
return 0;
}

Compilation warning:

gcc test.c -o test
test.c:24:34: warning: format specifies type 'unsigned int *' but the argument has type 'uint8_t *' (aka 'unsigned char *') [-Wformat]
    sscanf(hexstring, "%02x", &input[i]);
                       ~~~~   ^~~~~~~~~

`format specifies type 'unsigned int *' but the argument has type 'uint8_t *' (aka 'unsigned char *')` At what point in the above message things become unclear? — n. m. could be an AI, Oct 20 '19 at 13:44
@raj123 n.m is telling you that the warning tells you exactly what's wrong. You're using the wrong specifier. — klutt, Oct 20 '19 at 13:47
That I know its required unsigned int* however I am using uint8_t. I need a way to resolve it without modifying input buffer type as input buffer is hex byte so I am using uint8_t*. @klutt — raj123, Oct 20 '19 at 13:50
@raj123 Have you read the documentation for scanf format string? — klutt, Oct 20 '19 at 13:51
I just typecasted uint8*t to unsigned int* while pasing in scantf — raj123, Oct 20 '19 at 13:53
Well the obvious way is to use something other than your buffer. How about `int c; sscanf(hexstring, "%02x", &c);`? You now have something you can probably put in the buffer somehow. Another way is outlined in the answer. — n. m. could be an AI, Oct 20 '19 at 13:55
"I just typecasted uint8*t to unsigned int*". This is undefined behaviour. You cannot just throw casts at problems and hope they go away. — n. m. could be an AI, Oct 20 '19 at 13:56
Ok thanks got it .Answer given by @MikeCAT is the proper solution for that@n.m. — raj123, Oct 20 '19 at 13:57
The simpler solution, which the `scanf` documentation should have showed you, is `%hhx`. — Steve Summit, Oct 20 '19 at 14:03

score 3 · Accepted Answer · answered Oct 20 '19 at 13:49

The size of uint8_t is usually small than one of unsigned int, so your code seems happened to work.

You should use correct format specifer.

Add

#include <inttypes.h>

at the beginning of your code and change the format "%02x" to "%02" SCNx8.

The SCNx8 macro will be expanded to correct format specifier and connected to the "%02" part.

If unfortunately this is not supported in your environment, another way is to use a buffer with correct type to read the value.

   for (int i = 0; i < array_size; i++)
   {
       unsigned int buffer = 0;
       sscanf(hexstring, "%02x", &buffer);
       input[i] = (uint8_t)buffer;
       hexstring = hexstring + 2;
   }

A simpler solution is `%hhx`. – Steve Summit Oct 20 '19 at 14:02 — Steve Summit, Oct 20 '19 at 14:02

How to resolve GCC compilation [-Wformat] warning?

1 Answers1