[英]TypeScript type checking on type rather than instance
I want to be able to pass a type (rather than an instance of the type) as a parameter, but I want to enforce a rule where the type must extend a particular base type 我希望能够传递一个类型(而不是该类型的实例)作为参数,但是我想强制执行一个规则,其中该类型必须扩展特定的基本类型
Example 例
abstract class Shape {
}
class Circle extends Shape {
}
class Rectangle extends Shape {
}
class NotAShape {
}
class ShapeMangler {
public mangle(shape: Function): void {
var _shape = new shape();
// mangle the shape
}
}
var mangler = new ShapeMangler();
mangler.mangle(Circle); // should be allowed.
mangler.mangle(NotAShape); // should not be allowed.
Essentially I think I need to replace shape: Function
with something...else? 从本质上讲,我认为我需要替换
shape: Function
用其他东西代替shape: Function
?
Is this possible with TypeScript? TypeScript可以实现吗?
Note: TypeScript should also recognise that shape
has a default constructor. 注意:TypeScript还应该识别
shape
具有默认构造函数。 In C# I would do something like this... 在C#中,我会做这样的事情...
class ShapeMangler
{
public void Mangle<T>() where T : new(), Shape
{
Shape shape = Activator.CreateInstance<T>();
// mangle the shape
}
}
There are two options: 有两种选择:
class ShapeMangler {
public mangle<T extends typeof Shape>(shape: T): void {
// mangle the shape
}
}
Or 要么
class ShapeMangler {
public mangle<T extends Shape>(shape: { new(): T }): void {
// mangle the shape
}
}
But both of these will be fine with the compiler: 但是这两种方法对于编译器都可以:
mangler.mangle(Circle);
mangler.mangle(NotAShape);
With the example you posted because your classes are empty, and an empty object matches every other object in structure. 在您发布的示例中,因为您的类为空,并且空对象与结构中的所有其他对象匹配。
If you add a property, for example: 如果添加属性,例如:
abstract class Shape {
dummy: number;
}
Then: 然后:
mangler.mangle(NotAShape); // Error
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