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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("-------------------------------------------------------\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:

Flowchart: Add two numbers using call by reference

C Programming Code Editor:

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C Programming: Tips of the Day

Static variable inside of a function in C

The scope of variable is where the variable name can be seen. Here, x is visible only inside function foo().

The lifetime of a variable is the period over which it exists. If x were defined without the keyword static, the lifetime would be from the entry into foo() to the return from foo(); so it would be re-initialized to 5 on every call.

The keyword static acts to extend the lifetime of a variable to the lifetime of the programme; e.g. initialization occurs once and once only and then the variable retains its value - whatever it has come to be - over all future calls to foo().

Ref : https://bit.ly/3fOq7XP