How to write Program to swap two variables using function call by value?

Answer 1

//This program swaps the values in the variable using function containing reference arguments

#include<iostream.h>

void swap(int &iNum1, int &iNum2);

void main()

{

int iVar1, iVar2;

cout<<"Enter two numbers "<<endl;

cin>>iVar1;

cin>>iVar2;

swap(iVar1, iVar2);

cout<<"In main "<<iVar1<<" "<<iVar2<<endl;

}

void swap(int &iNum1, int &iNum2)

{

int iTemp;

iTemp = iNum1;

iNum1 = iNum2;

iNum2 = iTemp;

cout<<"In swap "<<iNum1<<" "<<iNum2<<endl;

}

Reference arguments are indicated by an ampersand (&) preceding the argument:

int &iNUm1;

the ampersand (&) indicates that iNum1 is an alias for iVar1 which is passed as an argument.

The function declaration must have an ampersand following the data type of the argument:

void swap(int &iNum1, int &iNum2)

The ampersand sign is not used during the function call:

swap(iVar1, iVar2);

The sample output is

Enter two numbers

12

24

In swap 24 12

In main 24 12

------------------------------------------------------------------

By Satish from here

/ * Program to Swap Two Numbers by Using Call By Reference Method * /

#include

main()

{

int i, j;

clrscr();

printf("Please Enter the First Number in A : ");

scanf("%d",&i);

printf("\nPlease Enter the Second Number in B : ");

scanf("%d",&j);

swapr(&i,&j); /* call by reference*/

printf("A is now in B : %d",i);

printf("B is now in A : %d",j);

}

/* call by reference function*/

swapr(int *x, int *y)

{

int t;

t=*x;

*x=*y;

*y=t;

}

Answer 2

#include <stdlib.h>

#include <stdio.h>

swap (int *a, int *b) {

*a ^= *b;

*b ^= *a;

*a ^= *b;

}

int main (int /*argc*/, char **/*argv*/) {

int a = 17;

int b = 375;

printf ("a: %d b: %d\n", a, b);

swap (&a, &b);

printf ("a: %d b: %d\n", a, b);

return 0;

}

Answer 3

#include

void main()

{

int a,b;

a=5;

b=6;

/* now adding b to a we have */;

a=b+a;

b=a-b; /* now here b will get value of a */;

a=a-b /*this will give a value of b */;

}

Answer 4

The only way to achieve this is to pass pointer variables. A pointer is a variable that holds a memory address and allows indirect access to that memory. When you pass a pointer to a function, the pointer is passed by value. But since the value of a pointer is a memory address, it is the same as pass by reference:

template<typename T>

void swap (T* a, T* b) {

T* temp = a;

a = b;

b = temp;

}

Answer 5

Swap by value achieves nothing since the values are simply copies of the values you actually want to swap. Swapping the copies does not swap the actual values. The only way to achieve this is by passing pointers. Pointers are always passed by value, but dereferencing the memory they point to makes them act like references.

void SwapByReference(int & a, int & b )

{

int c = a;

a = b;

b = c;

}

void SwapByValue(int * a, int * b )

{

int c = *a;

*a = *b;

*b = c;

}

Answer 6

for .NET C++ define the following template template void NET_Swap(T % A, T % B) { const T C = A; A = B; B = C; }

Answer 7

Only 'call-by-value' is possible in C, so you have to use pointers:

void swap (type *pa, type *pb) {

type tmp;

tmp = *pa;

*pa = *pb;

*pb = tmp;

}