const references

Learn safe and efficient passing with const references.

const references

If you do not want a parameter to be modified, use a const reference: void printValue(const int& n). This means n is read-only and is not copied.

A const reference is both safe and efficient. Safe, because the function cannot accidentally change the caller's variable. Efficient, because large objects are not copied. So when passing types like std::string or std::vector to a function, const T& is preferred.

Why const reference?

Passing a large object to a function by value is expensive, because the whole object is copied. Pass by reference avoids the copy but allows the function to modify the object.

To get both benefits, use a const reference: no copy, no modification. This is the standard way to pass large objects to a function in C++. For example void print(const std::string& s).

Exercise The printValue function

Define a void printValue(const int& n) function; in the body print n with std::cout. In main declare int x = 7; and call printValue(x).

Show example solution

#include <iostream>

void printValue(const int& n) {
    std::cout << n << "\n";
}

int main() {
    int x = 7;
    printValue(x);
    return 0;
}

Test yourself

const reference parameter

void printValue(____ int& n) { std::cout << n; }

Show answer

void printValue(const int& n) { std::cout << n; }

Which is a const reference declaration?
- const int p;
- int const p = x;
- const int& r = x;
- int& const r = x;
Show answer

const int& r = x; — const int& r = x; declares a read-only reference.
Match each expression to its meaning

const int& n

Const reference
printValue(x)

Call
void

Returns no value
n

Read-only

Next lesson: Dangling references

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