why can not we increment an array in similiar way as pointer in c?

Question

#include <stdio.h>

int main(){
    int arr[] = {1, 2, 3, 4};
    int *p;
    p = arr;
    printf("%d\n", *p);
    printf("%d\n", *arr);
    p++;
    printf("%d\n", *p);
}

This code outputs:

1
1
2

but when we add 2 lines as below:

 #include <stdio.h>

    int main(){
        int arr[] = {1, 2, 3, 4};
        int *p;
        p = arr;
        printf("%d\n", *p);
        printf("%d\n", *arr);
        p++;
        printf("%d\n", *p);
        arr++;
        printf("%d\n", *arr);
    }

This code outputs:

C:\Users\Hasnat\Desktop\test.c||In function 'main':|
C:\Users\Hasnat\Desktop\test.c|11|error: lvalue required as increment operand
=== Build failed: 1 error(s), 0 warning(s) (0 minute(s), 0 second(s)) ===

Why can not we increment an array in same way we increment pointer containing address of that array to get next element??

Because it's an array. That's kinda the answer. It just decays into a pointer if needed, but is a completely different animal. — Sami Kuhmonen, Dec 18 '15 at 13:05
You cant treat arrays and pointers similarly .You cannot use array as lvalue . — ameyCU, Dec 18 '15 at 13:08
Of course that doesn't explain why `void f(int arr[]) {arr++;}` compiles without error. — Mr Lister, Dec 18 '15 at 13:10
Why doesn't an apple taste like an orange? Both are spheres and grow on trees. — too honest for this site, Dec 18 '15 at 13:11
@MrLister: In the context of a function declaration, this `int arr[]` is equivalent to `int * arr`. — alk, Dec 18 '15 at 13:12
@MrLister: The standard does. It states it is converted to a pointer. http://port70.net/~nsz/c/c11/n1570.html#6.7.6.3p7 — too honest for this site, Dec 18 '15 at 13:13
You can't increment the pointer of an address. By doing `printf("%d\n", *arr + 1)` you can print the second element like you're trying to do, but that's it. — Talaria, Dec 18 '15 at 13:15

score 2 · Answer 1 · answered Dec 18 '15 at 13:14

To quote C11, chapter 6.5.2.4, Postfix increment and decrement operators

The operand of the postfix increment or decrement operator shall have atomic, qualified, or unqualified real or pointer type, and shall be a modifiable lvalue.

and the definition of modifiable lvalue is given in chapter 6.3.2.1 of the same standard, Lvalues, arrays, and function designators

An lvalue is an expression (with an object type other than void) that potentially designates an object; [...] A modifiable lvalue is an lvalue that does not have array type, does not have an incomplete type, does not have a constqualified type, and if it is a structure or union, does not have any member (including, recursively, any member or element of all contained aggregates or unions) with a constqualified type.

So, you cannot use ++ on an array. Simple.

http://www.gnu.org/software/gnu-c-manual/gnu-c-manual.html#Array-Subscripts **bold** The array subscript expression A[i] is defined as being identical to the expression (*((A)+(i))). This means that many uses of an array name are equivalent to a pointer expression. — Andre, Apr 04 '16 at 15:13

score 0 · Answer 2 · answered Dec 18 '15 at 13:12

Because C doesn't allow assigning to array. Arrays in expressions are automatically converted to a pointer pointing at the first element of the array unless they are used for the operand of sizeof operating or unary & (address) operator. (N1256 6.3.2.1)

As the error message shows, the operand arr is converted to a pointer pointing the first element of the array and the pointer isn't a lvalue, so it cannot be modified via increment operator.

why can not we increment an array in similiar way as pointer in c?

2 Answers2

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