Unexpected Pointer Behavior in C++: Why does variable point to a local variable's value when another pointer assignment is commented out?

I wrote this code:

#include <iostream>


int main()
{
    int x = 0;
    int *u, *k;
    
    {   
        int y = 9;
        u = &y;
    }
    
    int z = 11;
    k = &z;
    
    std::cout << "u = " << u << std::endl;
    std::cout << "k = " << k << std::endl;
    std::cout << "*u = " << *u << std::endl;
    std::cout << "*k = " << *k;

    return 0;
}

and as expected I received the output:

u = 0x7fff039b1440
k = 0x7fff039b1440
*u = 11
*k = 11

but by commenting out the line //k = &z; the output became like this:

u = 0x7ffebca2bc1c
k = 0
*u = 9

why does u point to the value 9 if the variable z should be located at this address?

>Solution :

as expected I received the output: […]

You have no valid justification for any particular expectation because your program has undefined behavior.

Here:

    {   
        int y = 9;
        u = &y;
    }

u is assigned the address of a variable whose lifetime ends when execution of the { ... } compound statement terminates. At that point, the value stored in u becomes indeterminate. Because you later read the (indeterminate) value of u, the behavior (of the whole execution of the whole program) is undefined.

That a different program (that also has undefined behavior) produces different output is not significant, no matter how closely related the programs’ sources are.