classic array vs std::array

Question

I try this simple program:

int* getElement(int arrayy[], int index) {
     return &arrayy[index];
}

int main() {

    int arrayy[4]{};

    *getElement(arrayy, 2)=50;

    std::cout << arrayy[2] << '\n';// here it prints 50 !

    return 0;
 }

getElement() will return the address of an element of my array through a pointer return, dereference it in the main and then change the value.

I want to do the same thing using std::array in the place of classic array.

int* getElement(std::array<int, 4> arrayy, int index) {
    return &arrayy[index];
}

int main() {


    std::array<int, 4> arrayy{};

    *getElement(arrayy, 2)=50;

    std::cout << arrayy[2] << '\n';//here it prints 0 !

    return 0;
}

in the first it prints 50 and in the second 0 !

does std::array pass by value in calls ?

Answer 1

In the case of passing a pointer by value, the pointed at values are the same in the function and the call site. So the first program does what you expect.

---

In the second program, it's different because std::array has value semantics, so passing it by value means that you are passing a copy of all the underlying memory as well.

Note that in your code in the second program, the result of 0 is not guaranteed. You are returning a pointer to a value inside the copied std::array which will die when the function returns. This invokes undefined behavior.

If you want to get the same effect as the first program, you need to pass the std::array by reference instead:

int* getElement(std::array<int, 4> &arrayy, int index) {
                                // ^  reference
    return &arrayy[index];
}