How to assign address of array to pointer? - c++

Question

#include <iostream>
int main() {
int arr[2] = {1, 2};
int *p;
// p = &arr; // Does not compile
p = &arr[0]; // compiles
std::cout << "&arr = " << &arr << std::endl;
std::cout << "&arr[0] = " << &arr[0] << std::endl;
}
When I try to print the address both print the same address. But when I try to assign p = &arr it does not compile. Is there something in standard that says something against assigning address of array to pointer. I just wanted to know the reason why p = &arr does not compile?
Clang actually says error: cannot initialize a variable of type 'int *' with an rvalue of type

Answer 1

p = &arr;
is a compiler error because the type of &arr is int (*)[2] -- pointer to an "array of 2 ints". Hence, it cannot be assigned to p, whose type is int*.
Even though &arr and &arr[0] evaluate to the same numerical value, they are different types.

Answer 2

So you have this:
arr[0] is the first item in memory
arr[1] is the second item in memory
This is equivalent to the following:
*((int*)arr + 0) is the first item in memory
*((int*)arr + 1) is the second item in memory
"Dereference" the pointer, this lets you access the memory you want instead of the number which represents it in memory (the pointer):
*((int*)arr + 0)
This is equivalent to:
arr[0]
If you wanted the address of any item you can do as shown here:
(int*)arr + Index
The address of the first item is the memory address of the start of the array, so the address of the array AND the first item is:
(int*)arr + 0 or just (int*)arr
Your code here gets the address of the first item, which is the same as the address of the array:
p = &arr[0]; // compiles
Whenever you place an ampersand ( & ) its equivalent to getting the address of, so here you are getting [the address of the address of the array], this is not the same as the [address of the array]
p = &arr; // Does not compile
The type of the address of the address of the array is:
int (*)[2];
Not:
int *p;
That is why it doesn't compile, the types don't match.
To help with type related errors like this one you can use typeid and decltype, in C++ this lets you print the name of the type in question.
Like this
#include <iostream>
using namespace std;
int main()
{
int arr[2] = {1, 2};
std::cout<< "typeid ptr_array is " << typeid(decltype(&arr)).name() << "\n";
std::cout<< "typeid ptr_item is " << typeid(decltype(&arr[0])).name() << "\n";
return 0;
}
The result is:
ptr_array is PA2_i (Pointer to Array of size 2 with int)
ptr_item is Pi (Pointer to int)
"P" from typeid means pointer, "A" means array.
You can play around yourself here:
https://wandbox.org/permlink/RNNxjTMSUnLqUo6q