I wrote the following code declaring an array of 5 pointers to int and then a pointer to the address of the mentioned array:
int* arr[5];
int** p=&arr;
At compilation with gcc, I get the following warning:
initialization from incompatible pointer type [-Wincompatible-pointer-types] int** p=&arr;
Why is int** p the wrong pointer type, isn't &arr a pointer to a pointer to an int?
Many thanks.
In the statement
int* arr[5];
arr is array of 5 int pointer. Let's say it's pointing to a which is array of 5 integer. It looks like
int a[5]= {1,2,3,4,5};
int* arr[5] = {a,a+1,a+2,a+3,a+4};
Now when you write
int** p=&arr;
which is wrong as p is of int** type and it should point to arr not to &arr.
isn't &arr a pointer to a pointer to an int? No, it's not, &arr is pointer to an array of pointers to int a[0] etc.
Correct way is
int **p= arr;
Now you can print the array a elements using p as
printf("%d %d\n",*p[0],*p[1]);
change int **p = &arr to either int **p = &arr[0] or int **p = arr
In the second option array name arr is converted to a pointer to its first element arr[0] (i.e. a pointer to a pointer to an int).
int* arr[5] is a pointer to a pointer to an int, so &arr is a pointer to a pointer to a pointer to an int.
int* arr[5] is not a pointer to a pointer. It is an array of pointers. So &arr is a pointer to an array of pointers to int.