How do pointer-to-pointers work in C? (and when might you use them?)

How do pointers-to-pointers work in C?
When might you use them?

Let's assume an 8 bit computer with 8 bit addresses (and thus only 256 bytes of memory). This is part of that memory (the numbers at the top are the addresses):

  54   55   56   57   58   59   60   61   62   63   64   65   66   67   68   69
+----+----+----+----+----+----+----+----+----+----+----+----+----+----+----+----+
|    | 58 |    |    | 63 |    | 55 |    |    | h  | e  | l  | l  | o  | \0 |    |
+----+----+----+----+----+----+----+----+----+----+----+----+----+----+----+----+

What you can see here, is that at address 63 the string "hello" starts. So in this case, if this is the only occurrence of "hello" in memory then,

const char *c = "hello";

... defines c to be a pointer to the (read-only) string "hello", and thus contains the value 63. c must itself be stored somewhere: in the example above at location 58. Of course we can not only point to characters, but also to other pointers. E.g.:

const char **cp = &c;

Now cp points to c, that is, it contains the address of c (which is 58). We can go even further. Consider:

const char ***cpp = &cp;

Now cpp stores the address of cp. So it has value 55 (based on the example above), and you guessed it: it is itself stored at address 60.

As to why one uses pointers to pointers:

The name of an array usually yields the address of its first element. So if the array contains elements of type t, a reference to the array has type t *. Now consider an array of arrays of type t: naturally a reference to this 2D array will have type (t *)* = t **, and is hence a pointer to a pointer.
Even though an array of strings sounds one-dimensional, it is in fact two-dimensional, since strings are character arrays. Hence: char **.
A function f will need to accept an argument of type t ** if it is to alter a variable of type t *.
Many other reasons that are too numerous to list here.

How do pointers to pointers work in C?

First a pointer is a variable, like any other variable, but that holds the address of a variable.

A pointer to a pointer is a variable, like any other variable, but that holds the address of a variable. That variable just happens to be a pointer.

When would you use them?

You can use them when you need to return a pointer to some memory on the heap, but not using the return value.

Example:

int getValueOf5(int *p)
{
  *p = 5;
  return 1;//success
}

int get1024HeapMemory(int **p)
{
  *p = malloc(1024);
  if(*p == 0)
    return -1;//error
  else 
    return 0;//success
}

And you call it like this:

int x;
getValueOf5(&x);//I want to fill the int varaible, so I pass it's address in
//At this point x holds 5

int *p;    
get1024HeapMemory(&p);//I want to fill the int* variable, so I pass it's address in
//At this point p holds a memory address where 1024 bytes of memory is allocated on the heap

There are other uses too, like the main() argument of every C program has a pointer to a pointer for argv, where each element holds an array of chars that are the command line options. You must be careful though when you use pointers of pointers to point to 2 dimensional arrays, it's better to use a pointer to a 2 dimensional array instead.

Why it's dangerous?

void test()
{
  double **a;
  int i1 = sizeof(a[0]);//i1 == 4 == sizeof(double*)

  double matrix[ROWS][COLUMNS];
  int i2 = sizeof(matrix[0]);//i2 == 240 == COLUMNS * sizeof(double)
}

Here is an example of a pointer to a 2 dimensional array done properly:

int (*myPointerTo2DimArray)[ROWS][COLUMNS]

You can't use a pointer to a 2 dimensional array though if you want to support a variable number of elements for the ROWS and COLUMNS. But when you know before hand you would use a 2 dimensional array.

How do pointer-to-pointers work in C? (and when might you use them?)

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