C++ LANGUAGE REFERENCES AS RETURN VALUE

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C++ LANGUAGE REFERENCES AS RETURN VALUE

Reference as Return Value 


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 
}