What is happening exactly...? Function - pass by reference

Question

I have a void function that outputs two variables. When I pass in a variable 'a' twice, the function just outputs 'b' from within the function. Why is this happening?

#include <iostream>

void f(int &a, int &b)
{
a = 4;
b = 10;
std::cout << a << b << std::endl;
}

int main()
{
int a = 6;

f(a, a);
//output: 1010
return 0;
}

Answer 1

You are passing a by reference to the function twice, so as a result you get a and b both referring your main function's 'a'. Now you change a(inside f) to 4. So now you're a(inside main) becomes 4 too. Since b is also referring to a(inside main), b too is changed to 4. So now,

a(inside f)=4
b(inside f)=4
a(inside main)=4

Now you make b=10, b refers to a(inside main) so you're main function a becomes 10 too. Your a(inside f) refers to the a(inside main) so it too changes to 10. So now the status of the variables is:

a(inside f)=10
b(inside f)=10
a(inside main)=10

You're printing a(inside f) and b(inside f) so you get the output 1010(without any space)

Answer 2

Happy to help 😃

Answer 3

That makes sense. :) Thank you very much Rishi

Answer 4

Manav Roy why are you asking this? Am just a college boy who knows a bit of Programming 😗

Answer 5

Manav Roy I learnt some Python in my high school days, it was in my 12th grade syllabus, forgot it right away. I learn a good amount of C++ last year, then i learn some Java 2 or 3 months back.

When did I start? I would say an year back

Answer 6

Manav Roy Learning cpp just for fun actually. My real job is being a teacher :)