优雅的解决方案复制，const和非const，getter？ [英] Elegant solution to duplicate, const and non-const, getters?

问题描述

你不要讨厌它

  class Foobar {
 public：
 Something& ; getSomething（int index）{
 //大，非平凡的代码块... 
 return something; 
} 
 
 const东西& getSomething（int index）const {
 //大，非平凡的代码块... 
 return something; 
} 
} 
  

其他一个，因为你不能从 const 版本（编译器错误）调用非 - const
需要从非 - const 调用 const 版本。 >

有没有一个真正优雅的解决方案，如果没有，最接近的是什么？

解决方案

我记得从一个有效的C ++书中，做到这一点的方法是通过从其他函数转出const来实现非const版本。

这不是特别漂亮，但它是安全的。因为成员函数调用它是非常量，对象本身是非常量，允许转出const。

  class Foo 
 {
 public：
 const int& get（）const 
 {
 //非平凡的工作
 return foo; 
} 
 
 int& get（）
 {
 return const_cast< int&>（static_cast< const Foo *>（this） - > get 
} 
}; 
  

Don't you hate it when you have

class Foobar {
public:
    Something& getSomething(int index) {
        // big, non-trivial chunk of code...
        return something;
    }

    const Something& getSomething(int index) const {
        // big, non-trivial chunk of code...
        return something;
    }
}

We can't implement either of this methods with the other one, because you can't call the non-const version from the const version (compiler error). A cast will be required to call the const version from the non-const one.

Is there a real elegant solution to this, if not, what is the closest to one?

解决方案

I recall from one of the Effective C++ books that the way to do it is to implement the non-const version by casting away the const from the other function.

It's not particularly pretty, but it is safe. Since the member function calling it is non-const, the object itself is non-const, and casting away the const is allowed.

class Foo
{
public:
    const int& get() const
    {
        //non-trivial work
        return foo;
    }

    int& get()
    {
        return const_cast<int&>(static_cast<const Foo*>(this)->get());
    }
};

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