how is the address changed when passing a pointer to a pointer in function-CodePudding

I am a beginner to c and I've written this example code to help myself understand pointer to pointer as well as call by reference. I understood pretty much everything except for myFunc: cout << &ptr << endl; // 0x7ffeefbff4d8

I was hoping to get some clarification on this line. Since we are treating the address passed in as the value of a pointer to pointer, aka int** ptr. How is the address of ptr generated? Is ptr somehow implicitly intialized? Thanks in advance!

void myFunc(int** ptr)
{
    cout << ptr << endl;  // 0x7ffeefbff520
    int b = 20;
    *ptr = &b;
    cout << &ptr << endl; // 0x7ffeefbff4d8
    cout << &b << endl;   // 0x7ffeefbff4d4
    cout << *ptr << endl; // 0x7ffeefbff4d4
    return;
}

int main()
{
    int *p;
    int a = 10;
    p = &a;
    cout << &a << endl; // 0x7ffeefbff51c
    cout << p << endl;  // 0x7ffeefbff51c
    cout << *p << endl; // 10
    cout << &p << endl; // 0x7ffeefbff520
    myFunc(&p);
    cout << *p << endl; // 20
    return 0;
}

CodePudding user response：

When you declare a function such as

void myFunc(int** ptr)

…you actually declare an argument which, even being a pointer to pointer to an int, remains passed by value. Consequently, it's treated like a local variable that would be declared at top of your function.

This variable is therefore declared in the stack, and this is the address you get when using "&".