Question

just brushing up on some C for a class and I’ve run across a little something that makes me scratch me head. For this code:

char * findString(const char * s){
/* Allocate space */    
char * ret = malloc(strlen(s) + 1);    
/* Copy characters */
char * n;
n = ret;
for ( ;*s != 0; s++)
    if (isLetter(*s))
        *n++ = *s;
*n = 0;   
/* return pointer to beginning of string */
return ret;

}

(We’re just assuming an isLetter that returns a 1/0).

The idea of the snippet is to take a string with a bunch of crap in it, and return a string that contains only the letters.

So, how does ‘ret’ work in this instance? I’m very confused by the returning of ‘ret’ when ‘n = ret’ is declared above the for loop and ‘ret’ never gets set to anything afterwards. Obviously I’m missing something here. Help!

-R. L.

Answer 1

both ret and n are pointers to the same block of memory. their ‘values’ are simply memory addresses — when you change *n, you change *ret, even though n and ret retain their original values.

//make n point to the beginning of the block of memory pointed
//to by ret
n = ret;

//iterate through the string which was passed to
//the function
for ( ;*s != 0; s++)
        //if the current character is a letter:
        if (isLetter(*s))
                //set the character pointed to by n to
                //the current character in the string, and then
                //make n point to the next one.
                *n++ = *s;

note that the loop increments n, and then after the loop sets the last character to 0 (to null terminate the string). Now, n points to the end of the string — but since ret was never changed it still points to the beginning of the memory that you malloced before the loop. When you return it, you’re returning a pointer to the new string which is the string you passed to the function, minus all non-letters.

Note that after this function returns, it is the caller’s responsibility to free() the memory allocated by the function, lest ye roam into memory leaks.