问题描述
在争取const正确性的同时,我经常发现自己在编写这样的代码
While striving for const-correctness, I often find myself writing code such as this
class Bar;
class Foo {
public:
const Bar* bar() const { /* code that gets a Bar somewhere */ }
Bar* bar() {
return const_cast< Bar* >(
static_cast< const Foo* >(this)->bar());
}
};
用于许多方法,例如 bar()
。编写这些手工调用const的非const方法很繁琐。此外,我感觉自己在重复自己–这让我感到难过。
for lots of methods like bar()
. Writing these non-const methods which call the const ones by hand is tedious; besides, I feel I am repeating myself – which makes me feel bad.
我该怎么做才能减轻这项任务? (不允许使用宏和代码生成器。)
What can I do to alleviate this task? (Macros and code generators are not allowed.)
编辑:除了litb的解决方案之外,我也喜欢我的解决方案。 :)
Besides litb's solution I also like my own. :)
推荐答案
另一种方法是编写一个调用该函数(使用CRTP)并从该函数继承的模板。
Another way could be to write a template that calls the function (using CRTP) and inherit from it.
template<typename D>
struct const_forward {
protected:
// forbid deletion through a base-class ptr
~const_forward() { }
template<typename R, R const*(D::*pf)()const>
R *use_const() {
return const_cast<R *>( (static_cast<D const*>(this)->*pf)() );
}
template<typename R, R const&(D::*pf)()const>
R &use_const() {
return const_cast<R &>( (static_cast<D const*>(this)->*pf)() );
}
};
class Bar;
class Foo : public const_forward<Foo> {
public:
const Bar* bar() const { /* code that gets a Bar somewhere */ }
Bar* bar() { return use_const<Bar, &Foo::bar>(); }
};
请注意,调用没有性能损失:由于成员指针作为模板参数传递,因此通话可以照常内联。
Note that the call has no performance lost: Since the member pointer is passed as a template parameter, the call can be inlined as usual.
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