Converting to hex with strcat?

**|HBO|** · 01-16-2010

Hi,

Code:

static char *tohex (const unsigned char *digest, int len) {
  int i;
  static char dest[64];
  for(i = 0; i < len; i++){
    strcat (dest, &digest[i]);
    //printf("%02x", digest[i]); 
  }
  dest[0] = '\0';
  //printf("\n");
return dest;
}

i am using md5. When i get md5 hash, i want it to be a hex value in order to use the digest in another function. But couldn't do it. Am i suppose to use sprintf instead? Why i am having a problem with strcat?

Thank you.

**MK27** · 01-16-2010

One issue here is that dest is an undefined local variable. You should set it to zero before you apply strcat to it, otherwise the end could be anywhere:

Code:

static char dest[64] = {0};

Beyond that, it is hard to tell what you are trying to do.

**|HBO|** · 01-16-2010

I just want to convert a md5 digest (such as ËâØº%úB¶]) to hex value (such as cbe2d8ba9c25fa42b65d9d133e8c4c32)

it's OK with

Code:

  for(i = 0; i < len; i++){
    printf("%02x", digest[i]); 
  }

But i want it to pass this hex value to another function. This why i am doing this. But i am stuck...

**MK27** · 01-16-2010

Ah. Strcat will definitely not work then, since you are converting those byte values to a string using "%x".

If printf works, then sprintf() will work. Unfortunately, it doesn't concatenate. BUT, you can always use a second pointer and move it forward 2 bytes at a time, since the hex representation will always be 2 bytes. IMPORTANT NOTE: hopefully you have taken that into consideration when sizing dest. It should be TWICE the size of digest, +1 for the null terminator. An unsigned char is one byte, but to represent that value in a string as hex you need two characters, get it? (16^2=256)

So something like:

Code:

   static char dest[64];
   char *ptr = dest;
  for(i = 0; i < len; i++){
    sprintf(ptr, "%02x", digest[i]);
     ptr+=2;
  }

BTW, why are you using a static char here?

**|HBO|** · 01-16-2010

Thank you very much. i appreciate your help. I owe you one

Yes, i'm aware of that. I was gonna do that, though

Thanks again for pointing that out.
Btw, it's my bad.

**|HBO|** · 04-09-2010

Originally Posted by MK27

Ah. Strcat will definitely not work then, since you are converting those byte values to a string using "%x".

If printf works, then sprintf() will work. Unfortunately, it doesn't concatenate. BUT, you can always use a second pointer and move it forward 2 bytes at a time, since the hex representation will always be 2 bytes. IMPORTANT NOTE: hopefully you have taken that into consideration when sizing dest. It should be TWICE the size of digest, +1 for the null terminator. An unsigned char is one byte, but to represent that value in a string as hex you need two characters, get it? (16^2=256)

So something like:

Code:

   static char dest[64];
   char *ptr = dest;
  for(i = 0; i < len; i++){
    sprintf(ptr, "%02x", digest[i]);
     ptr+=2;
  }

BTW, why are you using a static char here?

Long time no see. I am stuck with sth. How can i return a variable (char) from this function?

Code:

char c2h(const unsigned char *digest, int len)
{
   static char dest[64];
   char *ptr = dest;
  for(i = 0; i < len; i++){
    sprintf(ptr, "%02x", digest[i]);
     ptr+=2;
return dest;
  }

this is not working. Thank you.

**tabstop** · 04-09-2010

dest is not a char. If you want to return a pointer to char, then you need to change the return type of the function.

**MK27** · 04-09-2010

Originally Posted by |HBO|

Long time no see. I am stuck with sth. How can i return a variable (char) from this function?

You need to return a pointer to a string. The string cannot be local to the function, since it will cease to exist when the function returns. So you use malloc() to put it on the global heap:

Code:

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

char *eg() {
	char *p = malloc(12);
	strcpy(p, "hello world");
	return p;
}

int main() {
	char *test = eg();

	puts(test);
	free(test);

	return 0;
}

It's important to free() malloc'd memory when you are done with it.

Alternately, you can submit a pointer and fill it in:

Code:

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

void eg(char *p) {
	strcpy(p, "hello world");
}

int main() {
	char test[12];

	eg(test);
	puts(test);

	return 0;
}

**EVOEx** · 04-09-2010

MK27:
Another solution is to make the array static. So his first post had it right when it came to that.

**MK27** · 04-09-2010

Originally Posted by EVOEx

MK27:
Another solution is to make the array static. So his first post had it right when it came to that.

Ah. That's how standard lib static pointers work then! Didn't know you could do that, but I guess it makes sense.

If you need to keep the data from one such function call and make another one, however, you will have to copy that returned data.

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