问题描述
我有这样的事情:
template <typename T>
requires ::std::movable<T> || ::std::copyable<T> || ::std::is_void_v<T>
class ValueWrapper
{
// varIoUs functions to do stuff with value
};
class Value {
public:
Value(Value const &) = delete;
Value &operator =(Value const &) = delete;
Value(Value &&other) noexcept : x_{other.x_} { other.x_ = -1; }
Value &operator =(Value &&other) {
Value tmp{::std::move(other)};
x_ = tmp.x_;
tmp.x_ = -1;
return *this;
}
ValueWrapper<Value> do_something() const; // Generates error related to incomplete type. :-(
private:
int x_;
};
当然,它不起作用,因为在编译器看到 do_something
声明时,Value
是一个不完整的类型,并且无法测试不匹配的类型是否可以移动或可复制,且不作废。
我应该如何围绕这个进行设计?
我可以更改 ValueWrapper 以便单个成员函数需要一些东西,而不是让类需要一些模板参数。但是,这似乎很钝,因为该类的目的是包装可移动或可复制的东西。我可以将 do_something
移出 Value
类并使其成为一个免费函数。但这似乎过于严格,对于任何 Value
类的方法来说可能不是明智之举。
这里是否还有其他没有这些缺点的设计选择?
在我目前关注的特定非抽象案例中,ValueWrapper
恰好类似于 Boost expected
,因此用于表示错误或返回值。
编辑:到目前为止,我最喜欢的答案是使用 auto
,并且要求函数定义内联出现才能工作。如果你希望函数定义不内联,你可以做这堆体操。但是,这真的很奇怪。
#include <concepts>
#include <utility>
template <typename T>
requires ::std::movable<T> || ::std::copyable<T> || ::std::is_void_v<T>
class ValueWrapper
{
// varIoUs functions to do stuff with value
};
namespace priv_ {
// We can forward declare a class without its member functions.
// But we can't forward declare a function without being able to
// fully name all of its types.
class Silly;
}
class Value {
public:
Value() : x_{-1} { }
Value(Value const &) = delete;
Value &operator =(Value const &) = delete;
Value(Value &&other) noexcept : x_{other.x_} { other.x_ = -1; }
Value &operator =(Value &&other) {
Value tmp{::std::move(other)};
x_ = tmp.x_;
tmp.x_ = -1;
return *this;
}
// inline here is basically documenting that we intend to give an
// inline deFinition later. We might want to say what the actual
// return type is too.
auto inline do_something() const;
private:
int x_;
// And we can friend a forward declared class so all of its member
// functions are basically member functions of this class (and
// hence have unrestricted access to all member functions and
// variables).
friend class priv_::Silly;
};
namespace priv_ {
class Silly {
public:
// And finally,Now that the Value type is 'complete',we can
// use it as a parameter to the `ValueWrapper` template type.
static ValueWrapper<Value> p_do_something(Value const &v);
};
}
auto inline Value::do_something() const
{
// And Now that the declaration for `p_do_something` has been
// seen,we can call it.
return priv_::Silly::p_do_something(*this);
}
ValueWrapper<Value> foo()
{
Value v;
return v.do_something();
}
解决方法
在具有默认类型的模板方法中转换 do_something ()
怎么样?
我的意思是
template <typename U = Value>
ValueWrapper<U> do_something() const;
或者也
template <int...,typename U = Value>
ValueWrapper<U> do_something() const;
如果你想避免 do_something()
可以被称为解释类型。
该类型在成员函数定义中是完整的,因此我们可以使用推导出的返回类型:
auto do_something() const {
return ValueWrapper<Value>();
}