arrays and pointers arithmetic

Question

Suppose an array int a[10].

Why we can't do a=a+1 ? but the same is valid with a pointer variable.

int  *ptr = a;

ptr = a+1;

How are both scenarios seen practically?

we cannot move the location in memory of the 'a' array. However, we can change an external pointer to that array to point to the second element of the array — user3629249, Dec 10 '14 at 11:47

score 2 · Accepted Answer · answered Dec 10 '14 at 11:10

Because array locations are constant.

You can't change the value of a, since that represents the starting address of the array. Moving it doesn't make any sense.

With int *ptr; your variable ptr is just a single pointer and can of course be set to point to anywhere you like.

There's no contradiction here. It's a little like with functions, the name of a function evaluates to its address (called "a function pointer") but you can't assign to that either.

score -1 · Answer 2 · answered Dec 10 '14 at 12:21

-1

Because Array name is constant pointer pointing to its first element.You cannot change the value of constant variables,i,e what the const keyword is used for.

int a[10];      //here a (array variable is Const i,e you Cannot a=a+1)
int* const p=&a[0]; //here Also Same,Now p is Const i,e you Cannot p=p+1.

But Here:

int* pp=&a[0];//Here pp=pp+1 will work, Because pp is not Const.

answered Dec 10 '14 at 12:21

Mysterious Jack

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"Array name is constant pointer" - no, it's an array. The fact that it can't change doesn't mean it's `const`. – The Paramagnetic Croissant Dec 10 '14 at 12:27
Internally it act like a constant, It has to be Because it should always represent the Base Address of its First Element.Am Pretty Sure About it. – Mysterious Jack Dec 10 '14 at 13:03
1

I see you are sure about it, but you're still wrong. All I am saying is that it's not a `const` pointer. A `const` pointer would look like `int *const p;`. An array is an array and is not a pointer. It can't be modified, but **not** because it's a `const`, rather because it's an array. Neither `const`s nor array can be modified, but **that does not mean that something which cannot be modified is a `const`.** – The Paramagnetic Croissant Dec 10 '14 at 13:07

score -2 · Answer 3 · answered Dec 10 '14 at 11:46

-2

You can if the array is a function argument

#include <stdio.h>

int rval(int a[10]) {
    a = a + 1;
    return a[0];
    }

int main(){
    int a[10] = {0,1,2,3,4,5,6,7,8,9};
    printf ("%d\n", rval(a));
    return 0;
}

Program output

answered Dec 10 '14 at 11:46

Weather Vane

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of course, the function argument is NOT the array, but rather a copy of the pointer to the array. the pointer copy can be changed, but not the actual location of the beginning of the array in memory – user3629249 Dec 10 '14 at 11:49
Agreed, but it answers the question as asked. – Weather Vane Dec 10 '14 at 11:50
Hi,But when we passes an array to a function,it automatically converts to a pointer.Hence it'll tun fine.Right? – alisha Dec 10 '14 at 12:09
2

In this case, `a` is not an array, it's a pointer. – The Paramagnetic Croissant Dec 10 '14 at 12:26

arrays and pointers arithmetic

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