A C++ program can be made easier to read and maintain by using references rather than pointers. A C++ function can return a reference in a similar way as it returns a pointer.
When a function returns a reference, it returns an implicit pointer to its return value. This way, a function can be used on the left side of an assignment statement. For example, consider this simple program:
#include <iostream> #include <ctime> using namespace std; double vals[] = {10.1, 12.6, 33.1, 24.1, 50.0}; double& setValues( int i ) { return vals[i]; // return a reference to the ith element } // main function to call above defined function. int main () { cout << "Value before change" << endl; for ( int i = 0; i < 5; i++ ) { cout << "vals[" << i << "] = "; cout << vals[i] << endl; } setValues(1) = 20.23; // change 2nd element setValues(3) = 70.8; // change 4th element cout << "Value after change" << endl; for ( int i = 0; i < 5; i++ ) { cout << "vals[" << i << "] = "; cout << vals[i] << endl; } return 0; }
When the above code is compiled together and executed, it produces the following result:
Value before change vals[0] = 10.1 vals[1] = 12.6 vals[2] = 33.1 vals[3] = 24.1 vals[4] = 50 Value after change vals[0] = 10.1 vals[1] = 20.23 vals[2] = 33.1 vals[3] = 70.8 vals[4] = 50
When returning a reference, be careful that the object being referred to does not go out of scope. So it is not legal to return a reference to local var. But you can always return a reference on a static variable.
int& func() { int q; //! return q; // Compile time error static int x; return x; // Safe, x lives outside this scope }
[C++] Returning values by reference in C++
原文：http://www.cnblogs.com/tianhangzhang/p/4983694.html