问题描述
当你拥有它时你不讨厌它
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;
}
}
我们不能用另一个实现这两个方法，因为你不能从 const 版本调用非 const 版本(编译器错误).从非 const 版本调用 const 版本需要强制转换.
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.
有没有真正优雅的解决方案，如果没有，最接近的解决方案是什么?
Is there a real elegant solution to this, if not, what is the closest to one?
推荐答案
我记得在一本 Effective C++ 书籍中，这样做的方法是通过从其他函数中丢弃 const 来实现非 const 版本.
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.
它不是特别漂亮，但很安全.由于调用它的成员函数是非常量的，因此对象本身也是非常量的，并且允许丢弃 const.
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&>(const_cast<const Foo*>(this)->get());
}
};
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