Passing Value-Type Parameters (C# Programming Guide)
A value-type variable contains its data directly as opposed to a reference-type variable, which contains a reference to its data. Therefore, passing a value-type variable to a method means passing a copy of the variable to the method. Any changes to the parameter that take place inside the method have no affect on the original data stored in the variable. If you want the called method to change the value of the parameter, you have to pass it by reference, using the ref or out keyword. For simplicity, the following examples use ref.
Example
The following example demonstrates passing value-type parameters by value. The variable n is passed by value to the method SquareIt. Any changes that take place inside the method have no affect on the original value of the variable.
class PassingValByVal
{
static void SquareIt(int x)
// The parameter x is passed by value.
// Changes to x will not affect the original value of x.
{
x *= x;
System.Console.WriteLine("The value inside the method: {0}", x);
}
static void Main()
{
int n = 5;
System.Console.WriteLine("The value before calling the method: {0}", n);
SquareIt(n); // Passing the variable by value.
System.Console.WriteLine("The value after calling the method: {0}", n);
// Keep the console window open in debug mode.
System.Console.WriteLine("Press any key to exit.");
System.Console.ReadKey();
}
}
/* Output:
The value before calling the method: 5
The value inside the method: 25
The value after calling the method: 5
*/
The variable n, being a value type, contains its data, the value 5. When SquareIt is invoked, the contents of n are copied into the parameter x, which is squared inside the method. In Main, however, the value of n is the same, before and after calling the SquareIt method. In fact, the change that takes place inside the method only affects the local variable x.
The following example is the same as the previous example, except for passing the parameter using the ref keyword. The value of the parameter is changed after calling the method.
class PassingValByRef
{
static void SquareIt(ref int x)
// The parameter x is passed by reference.
// Changes to x will affect the original value of x.
{
x *= x;
System.Console.WriteLine("The value inside the method: {0}", x);
}
static void Main()
{
int n = 5;
System.Console.WriteLine("The value before calling the method: {0}", n);
SquareIt(ref n); // Passing the variable by reference.
System.Console.WriteLine("The value after calling the method: {0}", n);
// Keep the console window open in debug mode.
System.Console.WriteLine("Press any key to exit.");
System.Console.ReadKey();
}
}
/* Output:
The value before calling the method: 5
The value inside the method: 25
The value after calling the method: 25
*/
In this example, it is not the value of n that is passed; rather, a reference to n is passed. The parameter x is not an int; it is a reference to an int, in this case, a reference to n. Therefore, when x is squared inside the method, what actually gets squared is what x refers to: n.
A common example of changing the values of the passed parameters is the Swap method, where you pass two variables, x and y, and have the method swap their contents. You must pass the parameters to the Swap method by reference; otherwise, you will be dealing with a local copy of the parameters inside the method. The following is an example of the Swap method that uses reference parameters:
static void SwapByRef(ref int x, ref int y)
{
int temp = x;
x = y;
y = temp;
}
When you call this method, use the ref keyword in the call, like this:
static void Main()
{
int i = 2, j = 3;
System.Console.WriteLine("i = {0} j = {1}" , i, j);
SwapByRef (ref i, ref j);
System.Console.WriteLine("i = {0} j = {1}" , i, j);
// Keep the console window open in debug mode.
System.Console.WriteLine("Press any key to exit.");
System.Console.ReadKey();
}
/* Output:
i = 2 j = 3
i = 3 j = 2
*/