Incorrect Output after Function Call

Question

I expect the result is undefined or something, but the output is "10", why?

I suppose the memory is destroyed after the function gets called.

#include <iostream>
void f(int *p)
{
    int l = 20;
    int *k = &l;
    p = k;
}
int main()
{
    int i = 10;
    int *j = &i;
    f(j);
    std::cout << *j;
    return 0;
}

Answer 1

The result isn't undefined. You pass the pointer j by value, so you modify a copy of it inside the function. The original j is left unchaged, so the result is still 10.

Answer 2

The program is well formed as the local variable l is not accessed outside of f() as the pointer j is unchanged in main() (a copy of the pointer j is passed to f()).

Answer 3

Lets decompose your code.

You first assign 10 to i. Then make j point to the adress of i.

In f, you set the value of p, and not the value it points to, to k. As passed as a parameter of f(), p value is copied (the adress value).

In consequence, you never modify the value pointed by j, simply the local value of j in f(), that is why it stays equal to 10.

Answer 4

There is absolutely nothing wrong with the code. The parameter of f is not passed by reference, so f has no way to modify j, therefore j still points to i after f exits.

Answer 5

You are confused about passing by reference / passing by value. While you are passing a pointer, the pointer itself is a number that you pass to the f() function pointing to a chunk of memory; modifying the *p variable in f() (rather than the chunk of memory where *p points to) will have no effect in the main() function.

A pointer is just a variable that holds an address of memory. In itself, it is passed by value; only the variable to which the pointer points is passed by reference.