Initialize array of characters using pointers

Question

This problem is driving me crazy, I'm sure I'm missing something. I need to initialize an array of chars using only pointers. Below is the code I have so far:

    int p2(){
         /* Implements problem 2 of lab */

         // Create an array 
         char **s = (char**)malloc( 11 *sizeof(char));
         char *p = *s;
         char start ='A';

         while( p != s+10){
            *p = start;
            start++;
            p++;
         }

         return(0);
    }

The problem I'm having is I don't know how to address the characters inside of the array. I understand the base address of the array is **s, and the pointer to the first element is *s. What I don't understand is how to get to **s+10 (i.e. the end of the array).

Can anyone shine some light for me??? Please!

EDIT: Ok, looks like I misunderstood the question. I appears I need to create an array of strings (thus the char ** allocation). Then I need to loop through this array, and assign each string (i.e. char *) a value 15 chars long. Please let me know if I'm understanding this correctly:

char **strings ==> strings[0 ... n ] where each element is a pointer to a char (possibly an array). There for *string ==> strings[0], *(string+1) = strings[1], etc etc.

Am I close or way off?

Vyktor · Accepted Answer · 2012-01-22 10:03:09Z

4

char **s is 2 dimensional array of characters, or array of C strings if you want. If you want to use array of characters you should use:

char *string = (char*)malloc( 11 *sizeof(char));

If you really want to initialize array of strings, at first step you're initializing array of pointers, that's:

char **s = (char**)malloc( 11 *sizeof(char *));

Please note that I'm using char * inside sizeof. Than when you may use strings, but at first you must initialize each string.

s[0] = (char*) malloc( 15*size(char)); // This is actually string, 14 characters long (NULL terminated)
char *p = s[0]; // p is pointer to beginning of your string now

And there's two way how to address your string:

s[0][3] // 4th character of your string
p[3]

Or if you want to use just pointers:

char *p = *(s+0);
*(p+3); // 4th character
*((*(s+0))+3) // To do it "hardcore"

EDIT: added an example

When you have **char p and use p++ or p + 1, C increases memory address. *p operator tells compiler that you now want to work with data stored in memory, not with pointer. Therefor those two syntax do the same:

p[10] = 'a';
*(p+10) = 'a';

So if you want traverse both your dimensions, you should use:

for( int i = 0; i < 11; i++){
    char *p = *(s+i);
    for( int j = 0; j < 10; j++){
        *(p + j) = 'a'; // Say you wanna fill them with as
    }

    // You may also use this syntax:
    while( p < (*(s+i) + 10)){ // or use != instead of <
        *p = 'a';
        p++;
    }
}

edited Jan 22, 2012 at 10:03

answered Jan 21, 2012 at 23:51

Vyktor

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2 Comments

certifiedNoob Over a year ago

So how would you traverse the array (**s?). Would it be pCurr = *s; while(pCurr != *(s+10)) {... pCurr++}

Vyktor Over a year ago

@certifiedNoob I've added traversing examples, hope they are helpful

James M · Accepted Answer · 2012-01-21 23:48:34Z

2

I think you meant this:

     char *s = (char*) malloc(11 *sizeof(char));
     char *p = s;

In which case you'd address the characters with s[x]

answered Jan 21, 2012 at 23:48

James M

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Comments

sverre · Accepted Answer · 2012-01-21 23:49:44Z

1

First, you don't need a char ** unless you need an array of arrays.

Second, you can get to the end of the array with (*s)[10] or *((*s)+10)

Third, in C programming, don't cast the result of malloc()

answered Jan 21, 2012 at 23:49

sverre

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1 Comment

ryyker Over a year ago

...Thanks for the 3rd line in your post. Sadly, every other answer casts return of malloc

Drew Dormann · Accepted Answer · 2012-01-21 23:50:11Z

1

The code is falling apart here.

char **s = (char**)malloc( 11 *sizeof(char));

You're allocating enough memory for 11 chars, which sounds like what you want to do.

However, you're casting the address to those 11 chars to a (char**) as if you were allocating space for pointers, not chars.

answered Jan 21, 2012 at 23:50

Drew Dormann

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2 Comments

certifiedNoob Over a year ago

Yes, I need to allocate space for 10 pointers to characters ( a character array, aka string). Then I need to loop through the array and assign values to though pointers. How would I address the pointers I've created in **s?

Drew Dormann Over a year ago

@certifiedNoob: A character array/string doesn't need 10 pointers to it. One will do the job. And that's what malloc is returning. The one pointer to the beginning of the array.

Dan Tumaykin · Accepted Answer · 2012-01-21 23:57:53Z

1

Why should you use a double pointer for creating an array of chars?

char *s = (char *) malloc(10 * sizeof(char));
char *start = s; // save starting pointer

for(; s < start+10; s++)
    *s = 'a';

return 0;

If you allocate char ** essentially you're allocating an array of pointers to char(eg. array of strings). If you need char array, allocate char array. If you start working with pointers a design of stack and heap can be really helpful.

edited Jan 21, 2012 at 23:57

answered Jan 21, 2012 at 23:52

Dan Tumaykin

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