Question

Explain call by value and call by reference with example. 

Answer 1

Call by value:
• In this case the value of parameters is passed to the called function.
• In this case the actual parameters are not accessible by the called function.
• This is implemented by using a simple variable name.
• Hence the actual parameters remain unchanged in case of call by value. 

#include<stdio.h> 
#include<conio.h>
void main()
{
int a,b;
void swap (int a, int b);
clrscr();
printf("Enter two numbers:");
scanf("%d %d",&a,&b);
printf("The values of a and b in the main function before calling the swap function are %d and %d\n",a,b);
swap(a,b);
printf("The values of a and b in main function after calling the swap function are %d and %d\n",a,b);
getch();
}
void swap (int a, int b)
{
int temp;
temp=a;
a=b;
b=temp;
printf("The values of a and b in the swap function after swapping are %d and %d\n",a,b);
}

 

Call by reference:
• In this case the reference of variable is passed to the function by passing the address of parameters.
• In this case, since the address of the variables are available the called function can access the actual parameters.
• This are implemented by use of pointer variables.
• Hence the parameters can be altered if required in case of call by reference method 

#include<stdio.h>
#include<conio.h>
void main()
{ 
int a,b;
void swap (int *p1, int *p2);
clrscr();
printf("Enter two numbers:");
scanf("%d %d",&a,&b);
printf("The values of a and b in the main function before calling the swap function are %d and %d\n",a,b);
swap(&a,&b);
printf("The values of a and b in main function after calling the swap function are %d and %d\n",a,b);
getch();
}
void swap (int *p1, int *p2)
{
int temp;
temp=*p1;
*p1=*p2;
*p2=temp;
printf("The values of a and b in the swap function after swapping are %d and %d\n",*p1,*p2);
}