﻿ C Program: Add two numbers using call by reference - w3resource

# C Exercises: Add two numbers using call by reference

## C Pointer : Exercise-5 with Solution

Write a program in C to add numbers using call by reference.

Pictorial Presentation:

Sample Solution:

C Code:

``````#include <stdio.h>

int main()
{
long fno, sno, sum;

printf("\n\n Pointer : Add two numbers using call by reference:\n");
printf("-------------------------------------------------------\n");

printf(" Input the first number : ");
scanf("%ld", &fno);
printf(" Input the second  number : ");
scanf("%ld", &sno);
printf(" The sum of %ld and %ld  is %ld\n\n", fno, sno, sum);
return 0;
}
{
long sum;
sum = *n1 + *n2;
return sum;
}
```
```

Sample Output:

``` Pointer : Add two numbers using call by reference:
-------------------------------------------------------
Input the first number : 5
Input the second  number : 6
The sum of 5 and 6  is 11
```

Flowchart:

C Programming Code Editor:

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## C Programming: Tips of the Day

What does i = (i, ++i, 1) + 1; do?

In the expression (i, ++i, 1), the comma used is the comma operator

the comma operator (represented by the token ,) is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type).

Because it discards its first operand, it is generally only useful where the first operand has desirable side effects. If the side effect to the first operand does not takes place, then the compiler may generate warning about the expression with no effect.

So, in the above expression, the leftmost i will be evaluated and its value will be discarded. Then ++i will be evaluated and will increment i by 1 and again the value of the expression ++i will be discarded, but the side effect to i is permanent. Then 1 will be evaluated and the value of the expression will be 1.

It is equivalent to:

```i;          // Evaluate i and discard its value. This has no effect.
++i;        // Evaluate i and increment it by 1 and discard the value of expression ++i
i = 1 + 1;
```

Note that the above expression is perfectly valid and does not invoke undefined behavior because there is a sequence point between the evaluation of the left and right operands of the comma operator.

Ref : https://bit.ly/3saxONC

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