Etydhv

I want to convert strings to it's hex representation as string too (like hex dump programs), for example "abz" to "61627A"

char * strToHex( char * str )

 int length = strlen ( str );
 char * newStr = malloc( length * 2 );
 if ( !newStr ) shutDown ( "can't alloc memory" ) ;

 for ( int x = 0; x < length; x++)
 char y = str[ x ];
 sprintf ( newStr + x * 2, "%02X", y );
 
 return newStr;

ShutDown definition is omitted here, it is a function that calls perror and exit()

I designed strToHex to be used like

char * str = "abcdefghijklmnopqrstuvwxyz";
char * hex = strToHex(str);
printf("%sn",hex);
//outputs : 6162636465666768696A6B6C6D6E6F707172737475767778797A

Formatting

Most C formatting guides do not include spaces around the arguments to function calls, nor the expressions within an if-statement. For an example of a C style most C programmers would find acceptable, see OpenBSD's style(9) manual.

I choose to associate * with the variable name, rather than floating between the type and name. This disambiguates the following example:

int *a, b;

Here, a is a pointer to an integer, but b is only an integer. By moving the asterisk next to the name, it makes this clearer.

int length = strlen ( str );
char * newStr = malloc (length * 2 );
if ( !newStr) shutDown ( "can't allocate memory" ) ;

Becomes:

int const len = strlen(str);
char *const new_str = malloc(len * 2);

if (new_str == NULL) 
 shutDown("can't allocate memory");

Error checking

Rather than calling shutDown() and exit()ing the program, you should instead return an error value which can be checked by the caller of str_to_hex(). Because you return a pointer, you can return NULL to indicate an error occurred and the caller should check errno.

Likewise, on some systems your program can incorrectly exit when length == 0. If we look at the manual page for malloc(3):

Return Value

The malloc() and calloc() functions return a pointer to the allocated memory that is suitably aligned for any kind of variable. On error, these functions return NULL. NULL may also be returned by a successful call to malloc() with a size of zero, or by a successful call to calloc() with nmemb or size equal to zero.

So by returning NULL we account for the case where malloc(3) returns NULL on success.

if (new_str == NULL) 
 shutDown("can't alloc memory");

Becomes:

if (new_str == NULL) 
 return NULL;

If you choose, you can also check if str is NULL before calling strlen(). This is up to you, and it's not uncommon in C to ignore this case and leave it as user error.

Looping

Use the size_t type in your loop rather than int. size_t is guaranteed be wide enough to hold any array index, while int is not.

Using i rather than x is more common for looping variables.

The y variable isn't needed. You can simply use str[i] in its place.

In terms of performance there's likely a faster option than using sprintf(). You should look into strtol(3).

Conclusion

Here is the code I ended up with:

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

char *
str_to_hex(char const *const str)

 size_t const len = strlen(str);

 char *const new_str = malloc(len * 2);

 if (new_str == NULL) 
 return NULL;
 

 for (size_t i = 0; i < len; ++i) 
 sprintf(new_str + i * 2, "%02X", str[i]);
 

 return new_str;


int
main(void)

 char *str = "abz";
 char *hex = str_to_hex(str);

 if (hex == NULL && strlen(str) != 0) 
 /* error ... */
 

 printf("%sn",hex);

 free(hex);

Hope this helps!