Difference between the use of pointers for array and single data point?

Question

#include <iostream>
using namespace std;
const int MAX = 3;

int main ()
{
   int  var[MAX] = {10, 100, 200};
   int  *ptr;

   // let us have array address in pointer.
  ptr = var;
  for (int i = 0; i < MAX; i++)
   {
      cout << "Address of var[" << i << "] = ";
      cout << ptr << endl;

      cout << "Value of var[" << i << "] = ";
      cout << *ptr << endl;

      // point to the next location
      ptr++;
   }
   return 0;
}

I don't understand why this code should work as the correct syntax is

ptr=&var[0];

instead of

ptr=var;

if var would have been something like

int var=49;

then the my version of reasoning holds and earlier program fails. Is their any distinct concept of pointer for array and single data point?

The name of an array decays into the address of the first element of the array. — Thomas Matthews, Jun 16 '15 at 18:05

score 5 · Answer 1 · edited Jun 20 '20 at 09:12

I don't understand why this code should work as the correct syntax is
ptr=&var[0];
instead of
ptr=var;

The two ways of assigning a pointer are equivalent. In expressions that assign an array variable to a pointer C++ interprets the array name as a pointer to the initial element of the array, i.e. &var[0].

If var would have been something like
int var=49;
then the my version of reasoning holds and earlier program fails.

This is because C++ provides an implicit conversion to pointer from arrays, but not from scalar variables.

score 2 · Accepted Answer · answered Jun 16 '15 at 18:22

First, we should understand that pointers and arrays are handled the same in C or C++, at least once we've moved past the definition. Let's make sure we understand a few things. From the example you have described above, the following are equivalent (after ptr=var).

var[0], *var, ptr[0], ptr. All == 10.

In the case of:

ptr = &var[0];

That is assigning the address of the first element of var. Another way to say that is that &var is a pointer to a pointer to an int. We know that var is a pointer to an int (int *) and *var is an int.

If *var and var[0] are equivalent, then we're doing the equivalent of *(&var), which is just var.

Maybe to make this more clear, you could think about a different index than 0. var[1] is the same as (var + 1). It holds then that:

ptr = var + 1

is the same as

ptr = &var[1]

They're just two different ways to express the same thing.

In your example of int var=49, var is an int, not a pointer to an int (remember that an array is just a pointer, so pointer to an int is the same as an array of ints). You could express that as:

int var=49;
ptr = &var;

Then if you remove the loop (or set MAX=1) you'll get the result you're looking for.

Think of * as removing a level of addressing (pointers) and & as adding one. It's a little confusing because * does the opposite when you're defining them.

int **i;

Means a pointer to a pointer of an int. Then:

&i - pointer to a pointer to a pointer of an int. i - pointer to a pointer of an int. *i - pointer to an int **i - int i[0] - pointer of an int *i[0] - int

Pointers in C and C++ are simply another datatype. The question of "Is their any distinct concept of pointer for array and single data point?" is a bit flawed because in your example you have an array (which is equivalent to a pointer), not an array to a pointer. A single data point (in your second example int var=49) still has an address, but isn't an array, so it isn't a pointer type, it is an integer type. You could have easily described that as:

int var[1] = {49};

Which is an array (pointer) to a single int, which should not be confused with an int.

score -1 · Answer 3 · answered Jun 16 '15 at 18:10

-1

An array name holds the address of its first element in memory. i.e a pointer.

int array[2] = {1,2}; // "array" is a pointer to array[0]
int **p = &array;  //pointer to pointer to array[0]

answered Jun 16 '15 at 18:10

Acha Bill

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score -1 · Answer 4 · answered Jun 16 '15 at 18:15

The name of array is a reference to first index of array. The correct way of assigning a pointer to array is:

int *ptr = arrayName;

OR

int *ptr = &arrayName[0];

An other thing to keep in mind is that you can add offset to pointer to reference other elements of array. This will point to the first value of aray *ptr++; OR *ptr+=1;. Whereas doing this with variable pointer will yield unexpected results.

Difference between the use of pointers for array and single data point?

4 Answers4