Question

I am beginner at UNIX C development and need little bit of help. I have these two functions:

void edit_char(){

     int i;
     int length = strlen(expression);
     char *tmp = (char *) malloc((length+1)*sizeof(char));
     pom[0]='*';
     for(i=1;i<length+1;i++){
         tmp[i] = expression[i-1];
     }
     strcpy(expression,tmp);
     free((void *) tmp);
 }



 char *edit_char2(char *string){

     int i;
     int length = strlen(string);
     char *tmp = (char *) malloc((length+1)*sizeof(char));
     tmp[0]='/';
     for(i=1;i<length+1;i++){
         tmp[i] = string[i-1];
     }
     strcpy(string,tmp);
     free((void *) tmp);
     return string;
  }

edit_char() edits global variable char *expression – it puts a symbol “*” at the beginning. Second edit_char2() do almost the same, but instead of editing global variable, it edits string from argument.

First function works fine, the problem is with the tmp variable of second function. Malloc doesn’t return empty char array with size of (length+1). It returns “xd\372\267xd\372\267\020”.

What could cause this?

Answer 1

malloc just returns a pointer to the newly allocated memory, which is not to be considered “empty”. If you want to “empty” it – i.e, fill your memoy with NULLs, you need to do it manually need to use calloc or fill it manually, for instance with:

• bzero (deprecated)
• memset

EDIT: Also I think that since you are adding a character to each string, you shouldn’t use

newstring = malloc((strlen(string) + 1) * sizeof(char))

but instead

newstring = malloc((strlen(string) + 2) * sizeof(char))

To allocate space both for the new character AND the terminating \0 into account.

EDIT2: Which also means that your functions can’t work / are not safe ! You are trying to make both string and expression contain 1 more character than they were probably initially allocated for !