Why is there sometimes the & operator inside the parameters of a function?

Answer 1

void setA(int &a)
{
a = 6;
}

int main()
{
int a = 5;
cout << "a = " << a << "\n";
setA(a);
cout << "a = " << a << "\n";
}

Here is an example. When we pass a through setA, we pass it by reference, because the arguements of the function is &a (the memory address of a). We then set a to 6. And even if we didn't return a from setA, the a in the main evaluates to 6.

If we didn't include the & operator in the arguements, then a = 6 will only be true inside the function, whereas in main, a will still be 5.

Sorry if this is confusing, my explaination is a bit shit.

Answer 2

The andersand (&) operator returns the memory address of an object or a variable.

When used passing variables in a function e.g.

int func(int &a, int &b)

this is called pass by reference, and allows you to modify the original variable that was passed through. When using pass by value (without the &), you are creating temporary variables inside that function, which do not effect the original variables.

Answer 3

What do you mean by original variables?

Answer 4

It makes sense. Great explanation! So if I wanted to print 5 without removing the &, all I'd have to do is print *p in the cout?