C++ - Swap two numbers without using Temporary Variable
The value of two variables can be swapped without using any temporary variables. The method involves using operators like +, *, / and bitwise.
Example: Using + operator
In the below example, the + operator is used to swap the value of two variables x and y.
#include <iostream>
using namespace std;
static void swap(int, int);
static void swap(int x, int y) {
cout<<"Before Swap.\n";
cout<<"x = "<<x<<"\n";
cout<<"y = "<<y<<"\n";
//Swap technique
x = x + y;
y = x - y;
x = x - y;
cout<<"After Swap.\n";
cout<<"x = "<<x<<"\n";
cout<<"y = "<<y<<"\n";
}
int main() {
swap(10, 25);
return 0;
}
The above code will give the following output:
Before Swap. x = 10 y = 25 After Swap. x = 25 y = 10
Example: Using * operator
Like + operator, the * operator can also be used to swap the value of two variables x and y.
#include <iostream>
using namespace std;
static void swap(int, int);
static void swap(int x, int y) {
cout<<"Before Swap.\n";
cout<<"x = "<<x<<"\n";
cout<<"y = "<<y<<"\n";
//Swap technique
x = x * y;
y = x / y;
x = x / y;
cout<<"After Swap.\n";
cout<<"x = "<<x<<"\n";
cout<<"y = "<<y<<"\n";
}
int main() {
swap(10, 25);
return 0;
}
The above code will give the following output:
Before Swap. x = 10 y = 25 After Swap. x = 25 y = 10
Example: Using / operator
Smilarly / operator can also be used to swap the value of two variables x and y.
#include <iostream>
using namespace std;
static void swap(float, float);
static void swap(float x, float y) {
cout<<"Before Swap.\n";
cout<<"x = "<<x<<"\n";
cout<<"y = "<<y<<"\n";
//Swap technique
x = x / y;
y = x * y;
x = y / x;
cout<<"After Swap.\n";
cout<<"x = "<<x<<"\n";
cout<<"y = "<<y<<"\n";
}
int main() {
swap(10, 25);
return 0;
}
The above code will give the following output:
Before Swap. x = 10 y = 25 After Swap. x = 25 y = 10
Example: Using bitwise operator
The bitwise XOR (^) operator can also be used to swap the value of two variables x and y. It returns 1 when one of two bits at same position in both operands is 1, otherwise returns 0.
#include <iostream>
using namespace std;
static void swap(int, int);
static void swap(int x, int y) {
cout<<"Before Swap.\n";
cout<<"x = "<<x<<"\n";
cout<<"y = "<<y<<"\n";
//Swap technique
x = x ^ y;
y = x ^ y;
x = x ^ y;
cout<<"After Swap.\n";
cout<<"x = "<<x<<"\n";
cout<<"y = "<<y<<"\n";
}
int main() {
swap(10, 25);
return 0;
}
The above code will give the following output:
Before Swap. x = 10 y = 25 After Swap. x = 25 y = 10
Disadvantages of using above methods
- The multiplication and division based approaches fail if the value of one of the variable is 0.
- The addition based approach may fail due to arithmetic overflow. If x and y are too large, operation performed on operands may result into out of range integer.
