ClassS04CS141 | CPlusPlusDocs | recent changes | Preferences

A "reference" in C++ is a variable that refers to another variable. You can think of it as an alias.

http://www.parashift.com/c++-faq-lite/references.html is an excellent page on references.

Here is something home grown:

Assigning to a reference

Note that assigning to a reference changes the value of the variable to which the reference refers. To make sure you understand this, consider this program:
// tmp.cc
#include <iostream.h>
main() {
  int *A;
  A = new int[10];    // allocate array of 10 integers and point A to it

  if (! A) {
    cerr << "Out of memory!\n" << endl;

  for (int i = 0;  i < 10;  ++i)
    A[i] = i;

  int &B = A[3];

  B = 999;

  for (int i = 0;  i < 10;  ++i)
    cout << "A[" << i << "] = " << A[i] << endl;

  delete[] A;

Save this program as "tmp.cc", compile with "g++ -o tmp tmp.cc" and run the program. The output should be

A[0] = 0
A[1] = 1
A[2] = 2
A[3] = 999
A[4] = 4
A[5] = 5
A[6] = 6
A[7] = 7
A[8] = 8
A[9] = 9

Functions can return references

The second thing that I want you to understand is that a function can return a reference. For example:

// tmp2.cc
#include <iostream.h>

int & 
max_ref(int &a, int &b)  {
   if (a > b) return a;
   else return b;

main() {
  int x = 3, y = 4;

  max_ref(x,y) = 999;

  cout << "x = " << x << ", y = " << y << endl;


What do you think this will output? It should output "x = 3, y = 999".

Returning a reference is especially useful when implementing an array class, when you want to be able to do something like:

#include "Array.h" 

main() {
  Array<int> A;

  A[3] = 10;

To implement operator[] for the Array class so that A[3] can be assigned to as in this example, the operator[] member function will have to return a reference.

Be careful about the scope of the variable you reference

When you use a function to return a reference, make sure that the reference you return is still valid outside the function. For example, the following routine returns a reference to a variable that ceases to be defined outside the scope of the function. This can get you into all kinds of trouble:

// tmp3.cc
#include <iostream.h>

int & 
wrong()  {
  int a = 1;
  return a;

main() {

  wrong() = 10;  // this causes "10" to be written into a deallocated memory cell

  cout << wrong() << endl;  // this prints the contents of a deallocated memory cell


The compiler will compile it, but may warn you about the problem:

% make -k tmp3
g++     tmp3.cc   -o tmp3
tmp3.cc: In function `int& wrong()':
tmp3.cc:6: warning: reference to local variable `a' returned

and running the program will give you unpredictable results. Note that the exact same issue occurs with pointers to variables. For more information about this issue, see the CPlusPlusDocs page for information about variable scoping.

ClassS04CS141 | CPlusPlusDocs | recent changes | Preferences
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Last edited April 1, 2004 1:53 pm by Neal (diff)