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// Use multiple where clauses using System; // Gen has two type arguments and both have a where clause class Gen<T, V> where T : class where V : struct { T ob1; V ob2; public Gen(T t, V v) { ob1 = t; ob2 = v; } } class MultipleConstraintDemo { static void Main() { // This is OK because string is a class and // int is a value type Gen<string, int> obj = new Gen<string, int>("test", 11); // The next line is wrong because bool is not // a reference type Gen<bool, int> obj = new Gen<bool, int>(true, 11); } }

In this example, Gen takes two type arguments and both have a where clause Pay special attention to its declaration:

Sketch the graph of h( x) =

class Gen<T, V> where T : class where V : struct {

Notice the only thing that separates the first where clause from the second is whitespace No other punctuation is required or valid

Part I:

Using the pronouns you and your is one of the easiest ways to focus on your readers Consider the revised sentence discussed above:

When writing generic code, there will be times when the difference between value types and reference types is an issue One such situation occurs when you want to give a variable of a type parameter a default value For reference types, the default value is null For nonstruct value types, the default value is 0, or false for bool The default value for a struct is an object of that struct with all fields set to their defaults Thus, trouble occurs if you want to give a variable of a type parameter a default value What value would you use: null, 0, false, or something else For example, given a generic class called Test declared like this:

First note that the function is undefined at x = 1 1 We calculate that h ( x) = ( x+1) 2 Thus the graph is everywhere increasing (except at x = 1) 2 We also calculate that h ( x) = ( x+1) 3 Hence h > 0 and the graph is concave up when x < 1 Likewise h < 0 and the graph is concave down when x > 1 Finally, as x tends to 1 from the left we notice that h tends to + and as x tends to 1 from the right we see that h tends to Putting all this information together, we obtain the graph shown in Figure 37

class Test<T> { T obj; //

works in all classes The solution to this problem is to use another form of default, shown here: default(type) This is the operator form of default, and it produces a default value of the specified type, no matter what type is used Thus, continuing with the example, to assign obj a default value of type T, you would use this statement:

obj = default(T);

We notice that k ( x) = 3x 2 + 6x 9 = 3( x 1) ( x + 3) So the sign of k changes at x = 1 and x = 3 We conclude that k is positive when x < 3; k is negative when 3 < x < 1; k is positive when x > 3 Finally, k ( x) = 6x + 6 Thus the graph is concave down when x < 1 and the graph is concave up when x > 1 Putting all this information together, and using the facts that k( x) when x and k( x) + when x + , we obtain the graph shown in Figure 38

This will work for all type arguments, whether they are value or reference types Here is a short program that demonstrates default:

// Demonstrate the default operator using System; class MyClass { // } // Construct a default value of T class Test<T> { public T obj; public Test() { // The following statement would work only for reference types // obj = null; // can t use // The following statement will work only for numeric value types obj = 0; // can t use

18:

// This statement works for both reference and value types obj = default(T); // Works! }