我在 C# 中有这个构建器(自然地,这个例子被简化了):
class BusBuilder
{
Wheels mWheels = DefaultWheels;
int mRoute = 0;
public BusBuilder WithWheels(Wheels aWheels)
{
mWheels = aWheels;
return this;
}
public BusBuilder WithRoute(int aRoute)
{
mRoute = aRoute;
return this;
}
public Bus Build()
{
return new Bus { Wheels = mWheels, Route = mRoute };
}
}
它是这样使用的:
Bus bus =
new BusBuilder()
.WithWheels(someWheels)
.WithRoute(50)
.Build()
现在我想提取一个只包含一些方法的父类(super class):
class VehicleBuilder
{
Wheels mWheels = DefaultWheels;
public VehicleBuilder WithWheels(Wheels aWheels)
{
mWheels = aWheels;
return this;
}
}
class BusBuilder : VehicleBuilder
{
...
}
问题是现在我不会写了
Bus bus =
new BusBuilder()
.WithWheels(someWheels)
.WithRoute(50)
.Build()
因为 WithWheels
返回的是 VehicleBuilder
而不是 BusBuilder
,因此没有定义 WithRoute
方法。
你会如何设计这个?
最佳答案
构建器模式在继承方面有点痛苦。你可以这样做:
class VehicleBuilder<T> where T : VehicleBuilder<T>
{
private T @this;
protected VehicleBuilder()
{
// Or pass it in as a constructor parameter
@this = (T) this;
}
public T WithWheels(...)
{
return @this;
}
}
然后:
class BusBuilder : VehicleBuilder<BusBuilder>
{
...
}
此时,您的 WithWheels
方法仍将返回一个 BusBuilder
,因此您仍然可以调用 WithRoute
。
请注意,您还需要在每个派生类构建器中使用一个新的 Build
方法...
关于c# - builder 之间的继承——如何处理类型?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13949938/