Dangling references
- C++ course
- Beginner
- Lesson 75/116
- 5 min
Learn the dangling reference concept and safety.
Dangling references
A reference becomes invalid when the variable it is bound to goes out of scope; this is called a dangling reference. For example, if you return a reference to a local variable inside a function, the variable is destroyed when the function exits and the reference is left dangling.
Using a dangling reference is undefined behavior; the program may crash, give wrong results, or appear to work. So when writing a function that returns a reference, you must ensure the target outlives the caller.
Safe usage
To avoid dangling references, follow two simple principles. First: never return a reference to a local variable. Second: the reference's target must outlive the reference itself.
In this lesson's task, you will write the safe reference function increment(int& n). The caller's x lives before and after the function call, so the reference is safe. Reference parameters are safe in this regard.
Practice The increment function
Define a void increment(int& n) function; in the body write n += 1;. In main declare int x = 10;, call increment(x), and print x with std::cout.
Starter code
#include <iostream>
// Define the increment function
int main() {
// Declare x
// Add an increment call
// Print x
return 0;
}Sample solution
#include <iostream>
void increment(int& n) {
n += 1;
}
int main() {
int x = 10;
increment(x);
std::cout << x << "\n";
return 0;
}Expected output
11
Questions and answers
Review the lesson questions together with their answers. To test yourself, solve the lesson interactively.
1 What is a dangling reference?
- An uninitialized reference
- A non-const reference
- A reference that is not a pointer
- A reference to a variable that has gone out of scope Correct answer
Explanation: A dangling reference is bound to a variable that has gone out of scope.
2 Match each expression to its meaning
| Term | Meaning |
|---|---|
int& n |
Reference parameter |
Local variable |
Do not reference |
const int& |
Safe passing |
n += 1 |
Increment |
3 increment body
void increment(int& n) { ____ = n + 1; }void increment(int& n) { n = n + 1; }Explanation: n = n + 1; increases n by one.