5

I'm getting warning: assignment from incompatible pointer type [enabled by default] when i compile the following code:

int main() {
     int (*aptr) [5] = NULL;
     int arr[5] = {1,2,3,4,5};

     aptr = &arr[0];

     printf("aptr = %p\n arr = %p\n", aptr, &arr[0]);
     return 0;
}

I'm getting the correct output:

aptr = 0xbfcc2c64
arr = 0xbfcc2c64

But why am I getting the warning of incompatible pointer type?

banarun
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user2552690
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    possible duplicate of [Why does a pointer to array need to be cast before being passed as parameter to a function with array type argument?](http://stackoverflow.com/questions/17438175/why-does-a-pointer-to-array-need-to-be-cast-before-being-passed-as-parameter-to) – luser droog Jul 05 '13 at 07:02

2 Answers2

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You declared a pointer to the entire array. Why are you trying to make it point to the first element?

If you want to declare your aptr with int (*)[5] type, as in your example, and make it point to arr, then this is how you are supposed to set the pointer value

aptr = &arr;

What you have in your code now is an attempt to assign a int * value to a pointer of int (*)[5] type. These are different types, which is why you get the warning (which is a constraint violation, AKA error, actually).

AnT stands with Russia
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  • Thanks for the reply, the warning got removed. Just a small clarification, &arr is the address of the array. &arr[0] is address of 1st element in the array, conceptually both hold the same address.. isn't it? I tried to print the value &arr and &arr[0] which printed out, &arr = 0xbfa092e4 &arr[0] = 0xbfa092e4 Please excuse if is dumb question. – user2552690 Jul 05 '13 at 08:41
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    @user2552690: Yes, purely numerically the address of the entire array is always the same as the address of its first element. Yet the types of the pointers are different. As one consequence of that, pointer arithmetic works differently for such pointers, for one example. Try printing `&arr + 1` and `&arr[0] + 1` and you will see the difference. – AnT stands with Russia Jul 05 '13 at 08:52
  • One more query. I was expecting that *aptr would give me the value of 1st element in the array. But I require **aptr to get the value of first element in the array. why? aptr = 0xbfdbf7e4 *(aptr) = 0xbfdbf7e4 **(aptr) = 1 – user2552690 Jul 05 '13 at 09:23
  • @user2552690: Well, that would be the same mistake. `aptr` is a pointer to *the entire array*. This means that `*aptr` is the array itself, not the first element. In other words, `*aptr` is the same thing as `arr`. In order to get access to the first elelement you have to do `(*aptr)[0]` (cf. `arr[0]`) or `**aptr` (cf. `*arr`). There's nothing surprising here: in order to get that `aptr` you had to use `&` operator. Now, in order to get back to array you have to "compensate" for that `&` operator with an extra `*`. – AnT stands with Russia Jul 05 '13 at 17:20
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Array name itself give the base address it is not needed to use &arr.Moreover arr[0] represents the value at the first index.

hemanth reddy
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