C ++：引用和工厂(C++: references and factories)

编程入门行业动态更新时间:2024-10-11 23:27:35

我有这样的事情：

struct D { virtual void operator() {...}; } struct D1 : public D { virtual void operator() {...}; } struct D2 : public D { virtual void operator() {...}; } void foo(D &d) {...};

所以这很好，很好地控制了我的D的生命周期：

foo(D()); foo(D1()); foo(D2());

但我在多个地方选择我的D变体，所以我想要一个简单的工厂：

const D& make_D() { // BAD, returning reference to temporary if(is_tuesday()) return D1(); return D2(); }

我可以返回一个对象，而不是返回一个临时对象的引用，然后切片到基类。或者，我可以从我的工厂返回一个指针，但客户端必须删除它。其他更复杂的解决方案也会给客户带来更多负担。

有没有办法写出类似的东西

D& d = make_D(); foo(d);

（甚至foo(make_D()) ）？目标是将复杂性包含在各种D定义中和make_D()以便像foo()和那些调用这些函数的函数不需要担心它。

I have something like this:

struct D { virtual void operator() {...}; } struct D1 : public D { virtual void operator() {...}; } struct D2 : public D { virtual void operator() {...}; } void foo(D &d) {...};

And so this is fine, and nicely controls the life cycle of my D's:

foo(D()); foo(D1()); foo(D2());

But I choose my D variant in multiple places, so I want a simple factory:

const D& make_D() { // BAD, returning reference to temporary if(is_tuesday()) return D1(); return D2(); }

Instead of returning a reference to a temporary, I could return an object, but then I slice to the base class. Alternatively, I could return a pointer from my factory, but then the client has to delete it. Other, more complicated, solutions also impose more load on the client.

Is there a way to write something like

D& d = make_D(); foo(d);

(or even foo(make_D()))? The goal is to wrap the complexity in the various D definitions and in make_D() so that the functions like foo() and those who call those functions needn't worry about it.