Pointers to pointers vs. normal pointers

Question

The purpose of a pointer is to save the address of a specific variable. Then the memory structure of following code should look like:

int a = 5;
int *b = &a;         


        

Answer 1

Pointers are abstractions of memory addresses with additional type semantics, and in a language like C type matters.

First of all, there's no guarantee that int * and int ** have the same size or representation (on modern desktop architectures they do, but you can't rely on it being universally true).

Secondly, the type matters for pointer arithmetic. Given a pointer p of type T *, the expression p + 1 yields the address of the next object of type T. So, assume the following declarations:

char  *cp     = 0x1000;
short *sp     = 0x1000;  // assume 16-bit short
int   *ip     = 0x1000;  // assume 32-bit int
long  *lp     = 0x1000;  // assume 64-bit long

The expression cp + 1 gives us the address of the next char object, which would be 0x1001. The expression sp + 1 gives us the address of the next short object, which would be 0x1002. ip + 1 gives us 0x1004, and lp + 1 gives us 0x1008.

So, given

int a = 5;
int *b = &a;
int **c = &b;

b + 1 gives us the address of the next int, and c + 1 gives us the address of the next pointer to int.

Pointer-to-pointers are required if you want a function to write to a parameter of pointer type. Take the following code:

void foo( T *p )    
{
  *p = new_value(); // write new value to whatever p points to
}

void bar( void )
{
  T val;
  foo( &val );     // update contents of val
}

This is true for any type T. If we replace T with a pointer type P *, the code becomes

void foo( P **p )    
{
  *p = new_value(); // write new value to whatever p points to
}

void bar( void )
{
  P *val;
  foo( &val );     // update contents of val
}

The semantics are exactly the same, it's just the types that are different; the formal parameter p is always one more level of indirection than the variable val.