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:

C Exercises: Pictorial: Add two numbers using call by reference.

Sample Solution:

C Code:

#include <stdio.h>
long addTwoNumbers(long *, long *);
int main()
   long fno, sno, sum;
   printf("\n\n Pointer : Add two numbers using call by reference:\n"); 
   printf(" Input the first number : ");
   scanf("%ld", &fno);
   printf(" Input the second  number : ");
   scanf("%ld", &sno);   
   sum = addTwoNumbers(&fno, &sno);
   printf(" The sum of %ld and %ld  is %ld\n\n", fno, sno, sum);
   return 0;
long addTwoNumbers(long *n1, long *n2) 
   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: Add two numbers using call by reference

C Programming Code Editor:

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous: Write a program in C to add two numbers using pointers.
Next: Write a program in C to find the maximum number between two numbers using a pointer.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.

Follow us on Facebook and Twitter for latest update.

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

We are closing our Disqus commenting system for some maintenanace issues. You may write to us at reach[at]yahoo[dot]com or visit us at Facebook