Struct and function

Learn to pass structs to functions by value and by reference.

Passing a struct to a function

You can pass a struct variable to a function as a parameter. In pass-by-value, the function gets a copy of the struct; the original variable is not changed.

void printPoint(Point p) {
    cout << p.x << ", " << p.y;
}

Here printPoint takes a Point and prints its members. The call looks like printPoint(p);. Pass-by-value is fine for small structs; for large structs, references are used to avoid the copying cost.

Passing by reference

For large structs, pass-by-reference is used to avoid the copying cost. You add & next to the parameter type; this lets the function see the original struct directly.

void move(Point& p) {
    p.x += 1;
    p.y += 1;
}

Here p is a reference, so p.x += 1 changes the x member of the original struct. If you do not want the struct to change, use const Point&; this avoids copying and protects the data.

Exercise Write the printPoint function

Define a function void printPoint(Point p) that prints the members with cout << p.x << " - " << p.y << endl;. In main, declare Point p; p.x = 3; p.y = 4; and call printPoint(p);.

Show example solution

#include <iostream>
using namespace std;

struct Point {
    int x;
    int y;
};

void printPoint(Point p) {
    cout << p.x << " - " << p.y << endl;
}

int main() {
    Point p;
    p.x = 3;
    p.y = 4;
    printPoint(p);
    return 0;
}

Test yourself

What is used to pass a large struct without copying it?
- reference
- derived type
- constant
- pass by value
Show answer

reference — A reference lets you pass the struct without copying.

Fill in the blank: reference parameter

void move(Point____ p) {
    p.x += 1;
}

Show answer

void move(Point& p) {
    p.x += 1;
}

Match each passing style with its feature

by value

creates a copy
by reference

can change the original
const reference

no copy, protected

Next lesson: Array of structs

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