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The base class constraint enables you to specify a base class that a type argument must inherit A base class constraint serves two important purposes First, it lets you use the members of the base class specified by the constraint within the generic class For example, you can call a method or use a property of the base class Without a base class constraint, the compiler has no way to know what type of members a type argument might have By supplying a base class constraint, you are letting the compiler know that all type arguments will have the members defined by that base class The second purpose of a base class constraint is to ensure that only type arguments that support the specified base class are used This means that for any given base class constraint, the type argument must be either the base class, itself, or a class derived from that base class If you attempt to use a type argument that does not match or inherit the specified base class, a compile-time error will result

18:

d 2 ( x /x) using the quotient rule Of course x 2 /x = x, dx and you may calculate the derivative directly Observe that the two answers are the same The calculation confirms the validity of the quotient rule by way of an example Use a similar example to confirm the validity of the product rule

Here, T is the name of the type parameter, and base-class-name is the name of the base class Only one base class can be specified Here is a simple example that demonstrates the base class constraint mechanism:

// A simple demonstration of a base class constraint using System; class A { public void Hello() { ConsoleWriteLine("Hello"); } } // Class B inherits A class B : A { } // Class C does not inherit A class C { } // Because of the base class constraint, all type arguments // passed to Test must have A as a base class class Test<T> where T : A { T obj; public Test(T o) { obj = o; } public void SayHello() { // OK to call Hello() because it s declared // by the base class A objHello(); } } class BaseClassConstraintDemo { static void Main() { A a = new A(); B b = new B(); C c = new C(); // The following is valid because A is the specified base class Test<A> t1 = new Test<A>(a); t1SayHello(); // The following is valid because B inherits A Test<B> t2 = new Test<B>(b);

Part I:

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t2SayHello();

An important formula in the calculus relates the derivative of the inverse of a function to the derivative of the function itself The formula is [ f 1 ] (t) = 1 f ( f 1 (t)) ( )

// // } }

// The following is invalid because C does not inherit A Test<C> t3 = new Test<C>(c); // Error! t3SayHello(); // Error!

In this program, class A is inherited by B, but not by C Notice also that A declares a method called Hello( ) Next, notice that Test is a generic class that is declared like this:

We encourage you to apply the Chain Rule to the formula f ( f 1 (x)) = x to obtain a formal derivation of the formula ( )

class Test<T> where T : A {

The where clause stipulates that any type argument specified for T must have A as a base class Now notice that Test declares the method SayHello( ), shown next:

public void SayHello() { // OK to call Hello() because it s declared // by the base class A objHello(); }

This method calls Hello( ) on obj, which is a T object The key point is that the only reason that Hello( ) can be called is because the base class constraint requires that any type argument bound to T must be A or inherit A, and A declares Hello( ) Thus, any valid T will define Hello( ) If the base class constraint had not been used, the compiler would have no way of knowing that a method called Hello( ) could be called on an object of type T You can prove this for yourself by removing the where clause The program will no longer compile because the Hello( ) method will be unknown In addition to enabling access to members of the base class, the base class constraint enforces that only types that inherit the base class can be passed as type arguments This is why the following two lines are commented-out: