Linux C program - HEX how can I Prefix with 0x [closed] - c

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This program converts string to Hex.
#include <stdio.h>
#include <string.h>
int main(void) {
char text[] = "thank you";
int len = strlen(text);
char hex[100], string[50];
// Convert text to hex.
int i,j;
for ( i = 0, j = 0; i < len; i++, j+= 2) {
sprintf(hex + j, "%02X", text[i] );
printf("0x%X ", text[i] ); //this prints fine
}
printf("'%s' in hex is %s.\n", text, hex); //'thank you' in hex is 7468616e6b20796f75.
// Convert the hex back to a string.
len = strlen(hex);
for (i = 0, j = 0; j < len; i++, j+= 2) {
int val[1];
sscanf(hex + j, "%2x", val);
string[i] = val[0];
string[i + 1] = '\0';
}
printf("%s as a string is '%s'.\n", hex, string);
return 0;
}
But I need 0x74, 0x68, 0x61, 0x6E, 0x6B, 0x20, 0x79, 0x6F, 0x75
or assign to array
unsigned char key[] = {0x74, 0x68, 0x61, 0x6E, 0x6B, 0x20, 0x79, 0x6F, 0x75};
How to store to a string this hex value.

Answer 1

Use 0x in your sprintf() format, and increase the amount you increment j accordingly.
for ( i = 0, j = 0; i < len; i++, j+= 5) {
sprintf(hex + j, "0x%02X ", text[i] );
printf("0x%X ", text[i] );
}
Then you need to allow for this when scanning it. Skip over the first two characters, and then increment by 5.
There's also no need to make val an array, you just need a single int variable. And you can add the null terminator at the end of the loop, not each time through the loop.
for (i = 0, j = 2; j < len; i++, j+= 5) {
int val;
sscanf(hex + j, "%2x", &val);
string[i] = val;
}
string[i + 1] = '\0';
DEMO