Transferring variables with object reference. With or without &?

Question

#include <iostream>
class class1
{
public:int number = 2;
};

class class2
{
public:
    void method(class1 &obj)
    {
        obj.number;
        std::cout << obj.number;

    }
};

int main()
{
    class1 c1;
    class2 c2;
    c2.method(c1);
}

What is the difference between running the void method(class1 &obj) with or without &? Does it change use of memory?

I've heard it's nice to use & so you don't get a NullReferenceException. I can still access the variable without &, so should i use & in this dumb example?

Have you read any good C++ book? Have your learned how actual parameters are passed to functions and what is reference in C++? — 273K, Jan 30 '21 at 17:48
C++ doesn’t have a “NullReferenceException”. Wherever source you’re learning C++ from, consider getting a good textbook instead. — Sneftel, Jan 30 '21 at 18:56
@S.M. I've read some java, but not C++ books. Got a recommendation? — olemca, Jan 30 '21 at 19:17
https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list — Sneftel, Jan 30 '21 at 20:35

Pat. ANDRIA · Accepted Answer · 2021-01-30T19:09:34.400

To make it very simple, Pass by value means the actual value is passed on. Thus you need to copy this value to a variable to use it.

void method(class1 obj)

Pass by reference means a number (called an address) is passed on which defines where the value is stored. The address is passed and then no copy is done and the actual object is modified.

void method(class1 &obj)

Pass-by-references is considered to be more efficient than pass-by-value, because it does not copy the arguments to a new variable of the same type (as it uses the actual object).

Also beware as pass-by-value does not modify the real object (only the copy). A great example to look at for instance is the ̀swap two variables` as follows:

void swap(int a, int b)
{
   int t;

   t  = b;
   b = a;
   a = t;
}

If you call by value in a main function using `swap(x,y)̀, the inital value (before the call) and the final value (after the call) does not swap.

However, if you define the function instead by reference void swap(int& a, int& b) (nothing changes in the body of the function), you will see that calling swap(x,y), will indeed swap the two variables as the operation.

Thanks, my dad taught me this 15 minutes ago actually. – olemca Jan 30 '21 at 19:14 — olemca, Jan 30 '21 at 19:14
The point you made about swap was interesting! – olemca Jan 30 '21 at 19:26 — olemca, Jan 30 '21 at 19:26

Transferring variables with object reference. With or without &?

1 Answers1