In order to answer this question, we first need to understand what call-by-reference means. In call-by-reference, the function gets a reference to the actual variables passed in, not a copy of their values. This means that if the function changes the values of those variables, the changes are reflected in the original variables outside the function.

Let's assume that the swap function is defined as follows:

```cpp
void swap(int &x, int &y) {
int temp = x;
x = y;
y = temp;
}
```

This function swaps the values of the two variables passed to it.

Now, let's assume that `i` is an integer variable and `a` is an array of integers. Let's also assume that initially `i = 2` and `a = [1, 2, 3, 4, 5]`.

When we call `swap(i, a[i])`, we're passing `i` and `a[i]` (which is `a[2]` or `3`) to the swap function.

Here's what happens step by step:

1. Inside the swap function, `x` is a reference to `i` and `y` is a reference to `a[i]`. So `x = 2` and `y = 3`.

2. `temp` is assigned the value of `x`, so `temp = 2`.

3. `x` is assigned the value of `y`, so now `x = 3`. Since `x` is a reference to `i`, `i` also changes to `3`.

4. `y` is assigned the value of `temp`, so now `y = 2`. Since `y` is a reference to `a[i]` (which is now `a[3]` because `i` has changed), `a[3]` changes to `2`.

So, after the call to the swap function, `i = 3` and `a = [1, 2, 2, 4, 5]`.

Question

In order to answer this question, we first need to understand what call-by-reference means. In call-by-reference, the function gets a reference to the actual variables passed in, not a copy of their values. This means that if the function changes the values of those variables, the changes are reflected in the original variables outside the function.

Let's assume that the swap function is defined as follows:

```cpp
void swap(int &x, int &y) {
    int temp = x;
    x = y;
    y = temp;
}
```

This function swaps the values of the two variables passed to it.

Now, let's assume that `i` is an integer variable and `a` is an array of integers. Let's also assume that initially `i = 2` and `a = [1, 2, 3, 4, 5]`.

When we call `swap(i, a[i])`, we're passing `i` and `a[i]` (which is `a[2]` or `3`) to the swap function.

Here's what happens step by step:

1. Inside the swap function, `x` is a reference to `i` and `y` is a reference to `a[i]`. So `x = 2` and `y = 3`.

2. `temp` is assigned the value of `x`, so `temp = 2`.

3. `x` is assigned the value of `y`, so now `x = 3`. Since `x` is a reference to `i`, `i` also changes to `3`.

4. `y` is assigned the value of `temp`, so now `y = 2`. Since `y` is a reference to `a[i]` (which is now `a[3]` because `i` has changed), `a[3]` changes to `2`.

So, after the call to the swap function, `i = 3` and `a = [1, 2, 2, 4, 5]`.

Knowee AI · Accepted Answer