C# Out Parameter
C# Out Parameter
C# provides out keyword to pass arguments as out-type. It is like reference-type, except that it does not require variable to initialize before passing. We must use out keyword to pass argument as out-type. It is useful when we want a function to return multiple values.
C# Out Parameter Example 1
using System;
namespace OutParameter
{
class Program
{
// User defined function
public void Show(out int val) // Out parameter
{
int square = 5;
val = square;
val *= val; // Manipulating value
}
// Main function, execution entry point of the program
static void Main(string[] args)
{
int val = 50;
Program program = new Program(); // Creating Object
Console.WriteLine("Value before passing out variable " + val);
program.Show(out val); // Passing out argument
Console.WriteLine("Value after recieving the out variable " + val);
}
}
}
Output:
Value before passing out variable 50 Value after receiving the out variable 25
The following example demonstrates that how a function can return multiple values.
C# Out Parameter Example 2
using System;
namespace OutParameter
{
class Program
{
// User defined function
public void Show(out int a, out int b) // Out parameter
{
int square = 5;
a = square;
b = square;
// Manipulating value
a *= a;
b *= b;
}
// Main function, execution entry point of the program
static void Main(string[] args)
{
int val1 = 50, val2 = 100;
Program program = new Program(); // Creating Object
Console.WriteLine("Value before passing \n val1 = " + val1+" \n val2 = "+val2);
program.Show(out val1, out val2); // Passing out argument
Console.WriteLine("Value after passing \n val1 = " + val1 + " \n val2 = " + val2);
}
}
}
Output:
Value before passing val1 = 50 val2 = 100 Value after passing val1 = 25 val2 = 25
