Manipulating array outside of function using pointer? - c

Question

KEEP IN MIND I WROTE THIS CODE IN HERE TO SIMPLIFY MY QUESTION, PLEASE IGNORE MINOR MISTAKES IF THERE ARE ANY, MY PROBLEM FOCUSES ON MANIPULATING AN ARRAY OUTSIDE OF A FUNCTION BY USING IT'S POINTER.
I create a static array like this
char *array[size1][size2];
For the purposes of my example size1 = 3 and size2 = 6.
Then I use a function to pass the pointer of the array, and allocate strings to it.
int counter = 0;
void somefunction(char *a, char *b, int size1){
while(counter<size1){
//do some stuff to a
strcpy(b+counter, a);
counter++;
printf("Adding %s to pos %i RESULT: %s\n",a,counter,b+counter);
}
}
This works nicely and the output is correct (when outputting array b in the function), but it does not actually effect the array[] outside of the function. If I use
somefunction(a,array,size1);
The output will be correct, but the actual array outside of the function (array[]) will output gibberish random memory.
I output it like this:
for(int i=0;i<size1;i++){
printf("%s\n",array[i]);
}
So what am i doing wrong here? I thought the array when passed to a function decays into a pointer to the first element (and therefore that array), but that doesn't seem to be the case. What am I doing wrong? How do I manipulate the array outside of the function by passing it's pointer?
Thanks!

Answer 1

From the comments:
What Im trying to do is I calculate the length of the string (size2). So all of the strings contained will be of length 3 in this case, "ABC" for example. Size 1 is simply the amount of strings of size 3 that I need to store and iterate through. So size1=6 size2=3 means I want to have 6 strings accesable by ordered array, all of which are 3 chars in length. Once I compute size1 and size2 (I do this before creating the array) they do indeed stay constant.
In that case, you want to declare array as a 2D array of char:
char array[size1][size2]; // no *
Note that to store a string of length N, you need an N+1-element array to account for the 0 terminator. So if you're storing strings like "ABC" ({'A','B','C',0}), then size2 needs to be 4, not 3.
When you pass it to somefunction, the type of the expression array "decays" to "pointer to size2-element array of char" (char (*)[size2])1. Based on your description, it sounds like you declare array as a variable-length array (i.e., size1 and size2 are not known until runtime). If that's the case, then your function definition needs to look like
/**
* size2 must be declared in the argument list before char (*b)[size2]
* b may also be declared as char b[][size2]; it means the same thing
*/
void somefunction( char *a, size_t size2, char (*b)[size2], size_t size )
{
...
strcpy( b[counter], a );
...
}
and should be called as
somefunction( a, size2, array, size1 );
Except when it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T" and the value of the expression will be the address of the first element of the array. In your case, N is size1 and T is "size2-element array of char".

Answer 2

I believe you just need this:
char *array[SIZE];
Some function then looks like this:
int counter = 0;
void somefunction(char *a, char **b){
while(counter < SIZE){
//do some stuff to a
*(b+counter) = calloc(sizeof(char), strlen(a)+1);
strcpy(*(b+counter), a);
counter++;
printf("Adding %s to pos %i RESULT: %s\n",a,counter,b+counter);
}
}