这段与指针相关的代码有什么问题

Question

/* The structure of the Linked list Node is as follows:

struct Node
{
    int val;
    struct Node *next;

    Node(int data){
        val = data;
        next = NULL;
    }

}; 
*/

void intersection(Node **head1, Node **head2,Node **head3)
{

    cout<<*head1->val;
}

The above code in not working but when I take another pointer Node* h1=*head1; and then print its value its working fine. In both codes the value I want to print is same then why above code is wrong;

/* The structure of the Linked list Node is as follows:

struct Node
{
    int val;
    struct Node *next;

    Node(int data){
        val = data;
        next = NULL;
    }

}; 
*/

void intersection(Node **head1, Node **head2,Node **head3)
{

    Node* h1=*head1;
    cout<<h1->val;
}

Answer 1

In this code snippet

void intersection(Node **head1, Node **head2,Node **head3)
{

    cout<<*head1->val;
}

the expression

*head1->val

is equivalent to

*( head1->val )

(because the postfix operator -> has higher priority than the unary operator *) but the pointer head does not point to an object of the structure type. It points to another pointer, You have to write

( *head1 )->val

This is equivalent to the expression with the intermediate variable h1

Node* h1 = ( *head1 );
h1->val;

To make the difference more visible you can rewrite the expression of accessing the data member val the following way

( **head ).val

that is now the expression **head yields the lvalue of an object of the type struct Node .

Or using an intermediate variable like

Node *h1 = *head;
( *( h1 ) ).val
     ^
     |
   *head

Answer 2

Operator precedence puts -> before * . For illustration:

#include<iostream>

struct foo {
    int x;
};

int main() { 
    foo f;
    foo* fp = &f;
    foo** fpp = &fp;

    auto& xref = f.x;

    std::cout << &xref << "\n";
    //std::cout << *fpp->x;      // error
    std::cout << &(*fpp)->x;
}

The line marked as //error does not compile, beceause fpp has no member x . The other two lines print the same address, that of fx .