Print character array as hex in C

Question

I have a 2D array called char **str (allocated by malloc). Lets say str[0] has the string "hello". How would I print that hex?

I tried printf("%d\n", (unsigned char)strtol(str[0], NULL, 16)), but that doesn't print it out in hex.

Any help would be appreciated, thank you!

Solution 1

You are confused about the fuctionality of strtol. If you have a string that represents a number in hex, you can use strtol like you have:

char s[] = "ff2d";
int n = strtol(s, NULL, 16);
printf("Number: %d\n", n);

When you trying to print the characters of a string in hex, use %x format specifier for each character of the string.

char s[] = "Hello";
char* cp = s;
for ( ; *cp != '\0'; ++cp )
{
   printf("%02x", *cp);
}

Solution 2

Use %x flag to print hexadecimal integer

example.c

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) 
{
  char *string = "hello", *cursor;
  cursor = string;
  printf("string: %s\nhex: ", string);
  while(*cursor)
  {
    printf("%02x", *cursor);
    ++cursor;
  }
  printf("\n");
  return 0;
}

output

$ ./example 
string: hello
hex: 68656c6c6f

---

reference

• printf reference
• ascii table

Comments

[yelliver]

possible duplicate of Printing hexadecimal characters in C

[neby]

Thank you as well! The links really helped and always nice to see different ways to do the same thing. Also, the same question above what does the 02 do in %02x?

[amdixon]

02 is just zeropadding so in the event the hexadecimal integer was less than 2 bytes long eg. 9 then it would print 09.. not really required for the alpha characters as these are all in the two byte hexadecimal range

[not2qubit]

Can you explain the cursor thing? Also, this will give you compiler error unless you change *string to string[].