Why can't we directly assign arr to pointer to array

Question

I have an array of 5 integers. arr and &arr is same address. So why number 2 gives compilation error and and 3 works fine.

  int arr[5] = {1,2,3,4,5};
  1. int *p = arr;
  2. int (*p1)[5] = arr //Compilation Error
  3. int (*p1)[5] = &arr; //Works fine.
  
     arr = 0x61fdf0 and &arr= 0x61fdf0

Answer 1

The problem is that the initialized object and the initializer have different pointer types and there is no implicit conversion from one pointer type to another.

In this declaration

int (*p1)[5] = arr;

the initialized object has the type int ( * )[5] while the initializer has the type int * due to the implicit conversion of the array designator to a pointer to its first element.

You have to write

int (*p1)[5] = &arr

Or for example

int ( *p1 )[5] = reinterpret_cast<int ( * )[5]>( arr );

Here is a demonstrative program.

#include <iostream>

int main() 
{
    int arr[5] = {1,2,3,4,5};   
    
    int ( *p1 )[5] = reinterpret_cast<int ( * )[5]>( arr );

    std::cout << "sizeof( *p1 ) = " << sizeof( *p1 ) << '\n';
    
    return 0;
}

The program output is

sizeof( *p1 ) = 20

Objects can have equal values but different types.

Consider an another case.

Let's assume that you have an object of a structure type

struct A
{
    int x;
} a;

in this case &a and &a.x have the same values but the types of the expressions are different. You may not write for example

int *p = &a;

the compiler will issue an error. But you can write

int *p = ( int * )&a;

or

int *p = reinterpret_cast<int *>( &a );

Here is a demonstrative program.

#include <iostream>
#include <iomanip>

int main() 
{
    struct A
    {
        int x;
    } a = { 10 };
    
    std::cout << std::boolalpha << ( ( void * )&a.x == ( void * )&a ) << '\n';
    
    int *p = reinterpret_cast<int *>( &a );
    
    std::cout << "*p = " << *p << '\n';
    
    return 0;
}

The program output is

true
*p = 10

As for your one more question in a comment

Can you decode this in simple language to understand int** p = new int*[5];

then in this expression with the operator new

new int*[5]

there is allocated memory for an array with 5 elements of the type int *. The expression returns a pointer to the first element of the array. A pointer to an object of the type int * (the type of elements of the allocated array) will have the type int **.