为什么我不能投派生泛型类型立足非通用型（通过约束）？ [英] Why I cannot cast derived generic type to base non-generic type (through a constrain)?

问题描述

鉴于这种虚构的例子：

class NonGeneric
{
}

class Generic<T> : NonGeneric
    where T : NonGeneric
{
    T DoSomething()
    {
        return this; // **
    }
}

我希望它编译：通用＆LT; T＆GT;从非泛型导出和牛逼必须是一个派生类，它满足它的约束。

I'd expect it compiles: Generic<T> derives from NonGeneric and T must be a derived class so it satisfies its constrain.

我应该能够做到这一点：

I should be able to do this:

NonGeneric obj = new Generic<NonGeneric>();

那么就应该有这个指令没有问题：

Then there should be no problem in this instruction:

return this;

或者至少是这样的：

Or at least this:

return (T)this;

不幸的是它不工作，上面的例子中并没有错误编译：

Unfortunately it doesn't work and above example doesn't compile with error:

无法将类型非泛型＆LT; T＆GT; 来' T

我做错了什么，我看不到它，或者它只是不允许的？为什么这样？

I'm doing something wrong, and I can't see it, or it's just not allowed? Why this?

我会避免，如果可能的话，像一个什么解决办法我描述在这个岗位（反射，动态编译的方法等等）。我会避免动态对象太（设计决定，我不能改变的）。

I would avoid, if possible, any workaround like the one I describe in this post (Reflection, dynamic compiled methods and so on). I would avoid dynamic objects too (design decision, I can't change that).

推荐答案

您是否尝试过这个？

T DoSomething()
{
    return this as T;
}

这不隐式一样的 T 因此，有必要明确地将它转换为上述例。

this is not implicitly the same as T so there is a need to cast it explicitly as in the above example.

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