Functional Programming - Call By Reference

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In Call by Reference, the original value is changed because we pass reference address of arguments. The actual and formal arguments share the same address space, so any change of value inside the function is reflected inside as well as outside the function.

Call by Reference in C++

The following program shows how Call by Value works in C++ −

#include <iostream> 
using namespace std; 

void swap(int *a, int *b) {    
   int temp; 
   temp = *a; 
   *a = *b; 
   *b = temp; 
   cout<<"\n"<<"value of a inside the function: "<<*a; 
   cout<<"\n"<<"value of b inside the function: "<<*b; 
}  
int main() {     
   int a = 50, b = 75;   
   cout<<"\n"<<"value of a before sending to function: "<<a; 
   cout<<"\n"<<"value of b before sending to function: "<<b; 
   swap(&a, &b);  // passing value to function 
   cout<<"\n"<<"value of a after sending to function: "<<a; 
   cout<<"\n"<<"value of b after sending to function: "<<b; 
   return 0;   
}

It will produce the following output −

value of a before sending to function:  50 
value of b before sending to function:  75 
value of a inside the function:  75 
value of b inside the function: 50 
value of a after sending to function:  75 
value of b after sending to function:  50 

Call by Reference in Python

def swap(a,b): 
   t = a; 
   a = b; 
   b = t; 
   print "value of a inside the function: :",a 
   print "value of b inside the function: ",b 
   return(a,b) 
    
# Now we can call swap function 
a = 50 
b =75 
print "value of a before sending to function: ",a 
print "value of b before sending to function: ",b 
x = swap(a,b) 
print "value of a after sending to function: ", x[0] 
print "value of b after sending to function: ",x[1] 

It will produce the following output −

value of a before sending to function:  50 
value of b before sending to function:  75 
value of a inside the function:  75 
value of b inside the function:  50 
value of a after sending to function:  75 
value of b after sending to function:  50