Reference parameters

Learn pass-by-reference and how changes propagate to the caller.

Reference parameters

If you declare a function parameter as a reference, the function can modify the caller's variable directly. The syntax: void addOne(int& n). The & mark says the parameter is a reference, not a copy.

So a call addOne(x) changes x itself, not a copy. Pass by value does not propagate changes; pass by reference does. Reference parameters are also used to avoid copying large objects.

Pass by value versus by reference

In pass by value, the parameter is a copy of the argument; changes inside the function do not reach the caller. In pass by reference, the parameter is an alias bound to the argument itself; changes propagate at once.

void f(int n) is by value, void f(int& n) is by reference. The only visual difference is the & mark, but the behavior is completely different. Pass by reference is also used to return more than one value from a function.

Exercise The addOne function

Define a void addOne(int& n) function; in the body write n = n + 1;. In main declare int x = 5;, call addOne(x), and print x with std::cout.

Show example solution

#include <iostream>

void addOne(int& n) {
    n = n + 1;
}

int main() {
    int x = 5;
    addOne(x);
    std::cout << x << "\n";
    return 0;
}

Test yourself

What is the value of x after this code runs?
```
void addOne(int& n) {
    n = n + 1;
}
int main() {
    int x = 5;
    addOne(x);
    std::cout << x;
    return 0;
}
```
- 5
- 6
- 0
- Gives an error
Show answer

6 — addOne(x) passes by reference, so x increases: 5 → 6.
Match each expression to its meaning

int& n

Reference parameter
addOne(x)

Call
n = n + 1

Body
x

Argument

Reference parameter

void addOne(int____ n) { n = n + 1; }

Show answer

void addOne(int& n) { n = n + 1; }

Next lesson: const references

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