How do you write a program in C to swap two variables using a function?

Answer 1

#include<iostream>

void swap(int* x, int* x){

(*x)^=(*y)^=(*x)^=(*y);

}

int main()

{

int a, b;

a=10;

b=20;

std::cout<<"Before swap: a="<<a<<", b="<<b<<std::endl;

swap(&a,&b);

std::cout<<"After swap: a="<<a<<", b="<<b<<std::endl;

return(0);

}

Answer 2

To swap two variable using a function in C or C++, you need to pass the addresses, not the values, of the variables to the function.

swap (int *i, int *j) {

*i ^= *j;

*j ^= *i;

*i ^= *j;

}

example of call

int a = 12;

int b = 17;

swap (&a, &b);

Answer 3

C does not support move semantics thus in order to swap values we must copy them using a temporary variable.

int a = 42, b = 24;

// ...

// swap the values of a and b

int temp = a; a = b;

b = temp;

// a = 24, b = 42

Given the general usefulness of swapping, it makes sense to define a function. However, C only supports the pass by value semantic, so if we pass the values, only the formal argument values will be swapped, not the values of the actual arguments we passed. In order to swap the actual arguments we must pass their addresses (by value). This is also known as pass by reference even though C has no native reference type (C++ does have a native reference type).

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

int temp = *a;

*a = *b;

*b = temp;

}

Usage:

int x = 42, y = 24;

swap (&x, &y); // pass by reference

// x = 24, y = 42