Initialize array of characters using pointers - c

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?

Answer 1

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++;
}
}

Answer 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]

Answer 3

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()

Answer 4

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.

Answer 5

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.