qr code generator c# example is legal because B inherits A However, the second declaration in C#.NET

Generate QR Code ISO/IEC18004 in C#.NET is legal because B inherits A However, the second declaration

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// Gen<B, A> y = new Gen<B, A>();
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is illegal because A does not inherit B
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Using Multiple Constraints
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There can be more than one constraint associated with a type parameter When this is the case, use a comma-separated list of constraints In this list, the first constraint must be class or struct (if present) or the base class (if one is specified) It is illegal to specify both a class or struct constraint and a base class constraint Next in the list must be any interface constraints The new( ) constraint must be last For example, this is a valid declaration
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class Gen<T> where T : MyClass, IMyInterface, new() { //
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PART I PART I PART I
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In this case, T must be replaced by a type argument that inherits MyClass, implements IMyInterface, and has a parameterless constructor When using two or more type parameters, you can specify a constraint for each parameter by using a separate where clause Here is an example:
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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:
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:
The C# Language
Creating a Default Value of a Type Parameter
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 non-struct value types, the default value is 0 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, or something else For example, given a generic class called Test declared like this:
class Test<T> { T obj; //
if you want to give obj a default value, would you use
obj = null; // works only for reference types
obj = 0; // works only for numeric types and enums, but not structs
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);
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:
Generics
// This statement works for both reference and value types obj = default(T); // Works! }
PART I PART I PART I
// } class DefaultDemo { static void Main() { // Construct Test using a reference type Test<MyClass> x = new Test<MyClass>(); if(xobj == null) ConsoleWriteLine("xobj is null"); // Construct Test using a value type Test<int> y = new Test<int>(); if(yobj == 0) ConsoleWriteLine("yobj is 0"); } }
The output is shown here:
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