How to correctly convert a Hex String to Byte Array in C? -

i need convert string, containing hex values characters, byte array. although has been answered already here first answer, following error:

warning: iso c90 not support ‘hh’ gnu_scanf length modifier [-wformat] 

since not warnings, , omission of hh creates warning

warning: format ‘%x’ expects argument of type ‘unsigned int *’, argument 3 has type ‘unsigned char *’ [-wformat] 

my question is: how right? completion, post example code here again:

#include <stdio.h>  int main(int argc, char **argv) {     const char hexstring[] = "deadbeef10203040b00b1e50", *pos = hexstring;     unsigned char val[12];     size_t count = 0;       /* warning: no sanitization or error-checking whatsoever */     for(count = 0; count < sizeof(val)/sizeof(val[0]); count++) {         sscanf(pos, "%2hhx", &val[count]);         pos += 2 * sizeof(char);     }      printf("0x");     for(count = 0; count < sizeof(val)/sizeof(val[0]); count++)         printf("%02x", val[count]);     printf("\n");      return(0); } 

you can use strtol() instead.

simply replace line:

sscanf(pos, "%2hhx", &val[count]); 

with:

char buf[10]; sprintf(buf, "0x%c%c", pos[0], pos[1]); val[count] = strtol(buf, null, 0); 

update: can avoid using sprintf() using snippet instead:

char buf[5] = {"0", "x", pos[0], pos[1], 0}; val[count] = strtol(buf, null, 0);