C++中指针引用的生命周期

Question

I wrote some code that involves moving and changing variables in C++. Below is what I have wrote.

#include <iostream>

void six(int*& ptr) {
    int s = 6;
    ptr = &s;
}

int main() {
    int f = 5;
    int* ptr = &f;

    std::cout << *ptr << '\n';
    six(ptr);

    if (ptr) {
        std::cout << *ptr;
    }
    else {
        std::cout << "null\n";
    }

    return 0;
}

so this prints:

5
6

I tried another code, adding a single line:

#include <iostream>

void six(int*& ptr) {
    int s = 6;
    ptr = &s;
    free(&s); // added line
}

int main() {
    int f = 5;
    int* ptr = &f;

    std::cout << *ptr << '\n';
    six(ptr);

    if (ptr) {
        std::cout << *ptr;
    }
    else {
        std::cout << "null\n";
    }

    return 0;
}

Obviously, this gives an error after printing 5 because what the modified pointer is pointing is not available when called the second time.

However, I am confused at the first case. When calling six in the main function, variable s is not in the main scope, but the value itself still continues to remain in the memory to be referenced. Doesn't C++ automatically destroy variables and clean them when it goes out of the scope? Is this a memory leak?

Answer 1

The first case is not a memory leak, but an undefined behaviour because your variable go out of scope.

In this case you don't know when the memory will be cleaned(replaced) o reallocated.

So in some case the result can be correct but it's a pure question of luck.