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Passing Parameters by Reference to a Method
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By default, the CLR assumes that all method parameters are passed by value. When reference type objects are passed, the reference (or pointer) to the object is passed (by value) to the method. This means that the method can modify the object and the caller will see the change. For value type instances, a copy of the instance is passed to the method. This means that the method gets its own private copy of the value type and the instance in the caller isn t affected. Important In a method, you must know whether each parameter passed is a reference type or a value type because the code you write to manipulate the parameter could be markedly different. The CLR allows you to pass parameters by reference instead of by value. In C#, you do this by using the out and ref keywords. Both keywords tell the C# compiler to emit metadata indicating that this designated parameter is passed by reference, and the compiler uses this to generate code to pass the address of the parameter rather than the parameter itself. The difference between the two keywords has to do with which method is responsible for initializing the object being referred to. If a method s parameter is marked with out, the caller isn t expected to have initialized the object prior to calling the method. The called method can t read from the value, and the called method must write to the value before returning. If a method s parameter is marked 164
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with ref, the caller must initialize the parameter s value prior to calling the method. The called method can read from the value and/or write to the value. Reference and value types behave very differently with out and ref. Let s look at using out and ref with value types first:
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class App { static void Main() { Int32 x; SetVal(out x); Console.WriteLine(x); }
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// x doesn t have to be initialized. // Displays "10"
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static void SetVal(out Int32 v) { v = 10; // This method must initialize v. } }
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In this code, x is declared on the thread s stack. The address of x is then passed to SetVal. SetVal s v is a pointer to an Int32 value type. Inside SetVal, the Int32 that v points to is changed to 10. When SetVal returns, x has a value of 10 and "10" is displayed on the console. Using out with value types is efficient because it prevents instances of the value type s fields from being copied when making method calls. Now let s look at an example that uses ref instead of out:
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class App { static void Main() { Int32 x = 5; AddVal(ref x); // x must be initialized. Console.WriteLine(x); // Displays "15" } static void AddVal(ref Int32 v) { v += 10; // This method can use the initialize value in v. } }
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In this code, x is declared on the thread s stack and is initialized to 5. The address of x is then passed to AddVal. AddVal s v is a pointer to an Int32 value type. Inside AddVal, the Int32 that v points to is required to have a value already. So, AddVal can use the initial value in any expression it desires. AddVal can also change the value and the new value will be "returned" back to the caller. In this example, AddVal adds 10 to the initial value. When AddVal returns, Main s x will contain "15", which is what gets displayed in the console. To summarize, from an IL or a CLR perspective, out and ref do exactly the same thing: they both cause a pointer to the instance to be passed. The difference is that the compiler helps ensure that your code is correct. The following code that attempts to pass an uninitialized value to a method expecting a ref parameter produces a compilation error:
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class App { static void Main() { Int32 x;
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