CODEWITHSUNDEEP
c
the_naive
the_naive

Reputation: 3064

Parsing hexadecimal numbers into array

I have a variable char* etag="8a964ee2a5e88be344f36c22562a6486";.

How can I convert this into an array of byte, where byte is actually typedef unsigned char byte?

Upvotes: 1

Views: 3745

Answers (3)

Vijay
Vijay

Reputation: 67221

you probably use strtol

below is the example program:

/****************************************************************
 *
 * Purpose: To show examples of the strtol function.
 * Author:  M.J. Leslie
 * Date:    06-Nov-94
 *
 ****************************************************************/

#include <stdlib.h>

main()
{
  char num[10];

                /* Test a valid number      */
  strcpy(num,"13");

  printf("%s(Oct) is %i(Dec)\n", num, strtol(num, NULL,  8));
  printf("%s(Dec) is %i(Dec)\n", num, strtol(num, NULL, 10));
  printf("%s(hex) is %i(Dec)\n", num, strtol(num, NULL, 16));

  puts("----------------------------------");

                /* Test a slightly valid number
                 * Returns the same results as 
                 * above.           */
  strcpy(num, "13hzcd");

  printf("%s(Oct) is %i(Dec)\n", num, strtol(num, NULL,  8));
  printf("%s(Dec) is %i(Dec)\n", num, strtol(num, NULL, 10));
  printf("%s(hex) is %i(Dec)\n", num, strtol(num, NULL, 16));

  puts("----------------------------------");

                /* Test an invalid number
                 * Returns ZERO         */
  strcpy(num, "hzcd");

  printf("%s(Oct) is %i(Dec)\n", num, strtol(num, NULL,  8));
  printf("%s(Dec) is %i(Dec)\n", num, strtol(num, NULL, 10));
  printf("%s(hex) is %i(Dec)\n", num, strtol(num, NULL, 16));


  puts("----------------------------------");

                /* Test 0 base.
                 * This will look at the number 
                 * and decide the base for its self!
                 */
  strcpy(num, "13");
  printf("%s is %i(Dec)\n", num, strtol(num, NULL, 0));

  strcpy(num, "013");
  printf("%s is %i(Dec)\n", num, strtol(num, NULL, 0));

  strcpy(num, "0x13");
  printf("%s is %i(Dec)\n", num, strtol(num, NULL, 0));

}

/****************************************************************
 *
 * Results of the program:
 *
 *  13(Oct) is 11(Dec)
 *  13(Dec) is 13(Dec)
 *  13(hex) is 19(Dec)
 *  ----------------------------------
 *  13hzcd(Oct) is 11(Dec)
 *  13hzcd(Dec) is 13(Dec)
 *  13hzcd(hex) is 19(Dec)
 *  ----------------------------------
 *  hzcd(Oct) is 0(Dec)
 *  hzcd(Dec) is 0(Dec)
 *  hzcd(hex) is 0(Dec)
 *  ----------------------------------
 *  13 is 13(Dec)
 *  013 is 11(Dec)
 *  0x13 is 19(Dec)
 *
 ****************************************************************/

Upvotes: 2

Vaughn Cato
Vaughn Cato

Reputation: 64308

byte bytes[16];
int i=0;
for (;i<16;++i) {
  int value;
  sscanf(etag+2*i,"%02x",&value);
  bytes[i] = value;
}

Upvotes: 6

user529758
user529758

Reputation:

If I understand correctly, you want to make the 2-letter groups of hex chars to bytes, then store those bytes in your char array. Try this:

char *str = "long hex string";

int i;
unsigned char bytes[16];
char tmp[3] = { 0 };
for(i = 0; i < 16; i++)
{
    memcpy(tmp, str + 2 * i, 2);
    bytes[i] = (unsigned char)strtol(tmp, NULL, 16);
}

Upvotes: 5

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