Considere the following code in c :

#include <iostream>

using namespace std;

int main() {
    int x=2, y;
    int *p = &x;
    int **q = &p;
    std::cout << p << std::endl;
    std::cout << q << std::endl;
    std::cout << *p << std::endl;
    std::cout << x << std::endl;
    std::cout << *q << std::endl;

    *p = 8;
    *q = &y;

    std::cout << "--------------" << std::endl;

    std::cout << p << std::endl;
    std::cout << q << std::endl;
    std::cout << *p << std::endl;
    std::cout << x << std::endl;
    std::cout << *q << std::endl;

    return 0;
}

The output of code is (of course the list numbers is not the part of output):

1. 0x7fff568e52e0
2. 0x7fff568e52e8
3. 2
4. 2
5. 0x7fff568e52e0
6. '-----------'
7. 0x7fff568e52e4
8. 0x7fff568e52e8
9. 0
10. 8
11. 0x7fff568e52e4

Except for 7 and 9, all outputs were expected for me. I appreciate someone explaining them to me.

CodePudding user response：

The variable y was not initialized

int x=2, y;

So it has an indeterminate value.

As the pointer q points to the pointer p

int **q = &p;

then dereferencing the pointer q you get a reference to the pointer p. So this assignment statement

*q = &y;

in fact is equivalent to

p = &y;

That is after the assignment the pointer p contains the address of the variable y.

So this call

std::cout << p << std::endl;

now outputs the address of the variable y.

7. 0x7fff568e52e4

and this call

std::cout << *p << std::endl;

outputs the indeterminate value of y that happily is equal to 0.

9. 0