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?
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).
&arr + 1 and &arr[0] + 1 and you will see the difference.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 *.Array name itself give the base address it is not needed to use &arr.Moreover arr[0] represents the value at the first index.
