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C# Out ParameterC# 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 1using 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 2using 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
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