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C# program to swap two numbers
This post was published to provide the information and program in C# that are used to swap two numbers. I have created two C# programs, one without user input and the other with user input.
Let me clear up one thing: I am not going to use any third or temporary variable to swap two numbers.
Simple C# program to swap two numbers
Now let me create a simple C# program that swaps two numbers without using the third variable. Also, I am not allowing users to input any data through this program.
int numOne = 10; int numTwo = 20; Console.WriteLine("Before swap:"); Console.WriteLine("'numOne' = " + numOne + " 'numTwo' = " + numTwo); numOne = numOne + numTwo; numTwo = numOne - numTwo; numOne = numOne - numTwo; Console.WriteLine("\nAfter swap:"); Console.WriteLine("'numOne' = " + numOne + " 'numTwo' = " + numTwo);
The output produced by this C# example should exactly be
Before swap: 'numOne' = 10 'numTwo' = 20 After swap: 'numOne' = 20 'numTwo' = 10
C# program to swap two numbers entered by user
This C# program is created in a way to receive two numbers as input from the user at run-time of the program. Then those two numbers will be swapped and printed back on the output console.
int a, b; Console.WriteLine("Enter the first number: "); a = Convert.ToInt32(Console.ReadLine()); Console.WriteLine("Enter the second number: "); b = Convert.ToInt32(Console.ReadLine()); Console.WriteLine("\nBefore swap:"); Console.WriteLine("first number = " + a + " second number = " + b); a = a + b; b = a - b; a = a - b; Console.WriteLine("\nAfter swap:"); Console.WriteLine("first number = " + a + " second number = " + b);
The output produced by this C# program with user inputs 3 and 4 is shown in the snapshot given below.
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