It is a simple C++ program on “Call by Reference using Pointer“. It will perform a reference operation for the given variable. While compiling this call y reference program, input values are getting using the reference. This CPP program can be used for all kind of students for their lab exams and related tests. It may be useful for their exam purpose.

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CALL-BY-REFERENCE USING A POINTER PROGRAM

/* Title: Call by Reference
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*/
// Manually create a call-by-reference using a pointer.
#include <iostream.h>
#include <conio.h>
void neg(int *i);
int main()
{
clrscr();
int x;
x = 10;
cout << x << " negated is ";
neg(&x);
cout << x << "\n";
getch();
return 0;
}
void neg(int *i)
{
*i = -*i;
}
/*
In this program, neg() takes as a parameter a pointer to the integer whose sign it will reverse. Therefore, neg() must be explicitly called with the address of x. Further,inside neg() the * operator must be used to access the variable pointed to by i.
This is how you generate a "manual" call-by-reference in C++, and it is the only way to obtain a call-by-reference using the C subset. Fortunately, in C++ you can automate this feature by using a reference parameter.
To create a reference parameter, precede the parameter's name with an &. For example, here is how to declare neg() with i declared as a reference parameter:
void neg(int &i);
For all practical purposes, this causes i to become another name for whatever argument neg() is called with. Any operations that are applied to i actually affect the calling argument. In technical terms, i is an implicit pointer that automatically refers to the argument used in the call to neg() . Once "i" has been made into a reference, it is no longer necessary (or even legal) to apply the * operator. Instead, each time i is used, it
is implicitly a reference to the argument and any changes made to i affect the argument. Further, when calling neg(), it is no longer necessary (or legal) to precede the argument's name with the & operator. Instead, the compiler does this automatically.
*/