指向成员结构的指针如何工作？

Question

I am beginner in c++ and I want to question about how pointer to member struct work. Here I have code:

struct Node
{
    int data;
    Node* next;     // next variable store address of Node.

    void Print()
    {
        cout <<"Value in next variable is: " <<&next;
    }
};

int main()
{
    Node Sample;
    sample.data = 5;
    cout <<"Data is: " <<sample.data <<"\n";
    sample.Print();
    cout <<"Address of Sample is: " <<&Sample <<"\n";
}

Output:
Data is: 5
Value in next variable is: 0xbf9b6758
Address of Sample is: 0xbf9b6754

In this case pointer variable next has different address with &Sample . How this work? Why they have different address?

Answer 1

You are not printing the value of next. You take the address of next (by printing &next ). The address of the member next is the address is simply ((char *)&Sample) + offsetof(Node, next) . Ie the address of Sample plus the position of the member relative to the start of the struct.

The value of next could be printed by leaving out the & operator: cout << next , however you never set the value of next to begin with, so its value will be arbitrary.

Answer 2

Answer to query:

Value in next variable is: 0xbf9b6758 
Address of Sample is: 0xbf9b6754

These 4 steps learn you what is happening arround?

step 1: firstly, structure var sample is declared.

step 2: As sample is created into memory ,members of sample is initialized with default values.

step 3: becoz next is a pointer variable so, it will also initialized with some garbage value( ie next unused address in Ram).

step 4: next pointer holds the address - 0xbf9b6758 because,

"size of sample variable is = sum of sizes of each member of struct node"

= size of data + size of next

= 2 + 2 => 4

KeyPoint --> Size of next is 2 byte because Addresses in memory are of unsigned integer format(datatype)

So, next variable is assigned the different address from sample variable.