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

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

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 **.