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Returning values by reference in C++
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;
}
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When the above code is compiled together and executed, it produces 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
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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
}
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