Why does call-by-value example not modify input parameter?

Question

In the following call-by-value example, I cannot understand why this code is not changing the value of the 5 to a 6.

Line 11 calls the function changeValue which ha

Answer 1

The behavior being observed currently is because passing by value means a copy of value (new integer with value of value) is actually passed to the function.

You have to pass by reference. For that the changeValue function will look like this:

void changeValue(int& value)

Rest of the code will remain the same.

Passing a variable by reference means the same int value declared in main is passed to the changeValue function.

Alternatively, you can pass a pointer to value to the changeValue function. That will however, require changes to how you call the function also.

int main()
{
    int value = 5;

    changeValue(&value);

    ...

    return 0;
}

void changeValue(int* value)
{
    *value = 6;
}