提问人：prestokeys 提问时间：11/26/2017 最后编辑：Jodocusprestokeys 更新时间：11/26/2017 访问量：514
具有多重继承的复制赋值运算符
Copy assignment operator with multiple inheritance
问：
我下面的复制构造函数工作正常，但我不明白我的复制分配运算符出了什么问题。
#include <iostream>
template <typename... Ts> class foo;
template <typename Last>
class foo<Last> {
Last last;
public:
foo (Last r) : last(r) { }
foo() = default;
foo (const foo& other) : last(other.last) { }
foo& operator= (const foo& other) {
last = other.last;
return *this;
}
};
template <typename First, typename... Rest>
class foo<First, Rest...> : public foo<Rest...> {
First first;
public:
foo (First f, Rest... rest) : foo<Rest...>(rest...), first(f) { }
foo() = default;
foo (const foo& other) : foo<Rest...>(other), first(other.first) { std::cout << "[Copy constructor called]\n"; }
foo& operator= (const foo& other) { // Copy assignment operator
if (&other == this)
return *this;
first = other.first;
return foo<Rest...>::operator= (other);
}
};
int main() {
const foo<int, char, bool> a(4, 'c', true);
foo<int, char, bool> b = a; // Copy constructor works fine.
foo<int, char, bool> c;
// c = a; // Won't compile.
}
错误信息：
error: invalid initialization of reference of type 'foo<int, char, bool>&' from expression of type 'foo<char, bool>'
return foo<Rest...>::operator= (other);
^
有人可以在这里指出问题吗？
答：
8赞
StoryTeller - Unslander Monica
11/26/2017
#1
您的退货声明
return foo<Rest...>::operator= (other);
返回 a（这是用于定义的引用类型）。但它是从应该返回 .foo<Rest...>operator=foo<First, Rest...>&
从本质上讲，您返回一个需要引用的地方。引用根本不会绑定。BaseDerived&
幸运的是，修复很简单：不要返回 的结果，而是返回。foo<Rest...>::operator=*this
foo& operator= (const foo& other) { // Copy assignment operator
if (&other == this)
return *this;
first = other.first;
foo<Rest...>::operator= (other);
return *this;
}
评论
0赞
prestokeys
11/26/2017
@讲故事的人 是的，确实如此。作为奖励，它被多次调用，但冗余，对吧（它只需要调用一次）？但是没有办法避免这种情况吗？if (&other == this) return *this;
0赞
StoryTeller - Unslander Monica
11/26/2017
@prestokeys - 您始终可以委托给另一个不检查（并且以递归方式实现）的命名成员。operator=
0赞
prestokeys
11/26/2017
@StoryTeller 答案已接受，我实施了您的想法以在单独的答案中对其进行优化。谢谢。
1赞
Nir Friedman
11/26/2017
@prestokeys 实际上，您可能并不想以递归方式实现任何内容。请改用包扩展。在这些情况下可以节省大量的模板膨胀。
1赞
mp31415
11/26/2017
#2
看起来派生类中的返回不正确：operator=
return foo<Rest...>::operator= (other);
它返回基类，而它应该是 .将其更改为*this
this -> foo<Rest...>::operator= (other);
return *this;
0赞
prestokeys
11/26/2017
#3
感谢 StoryTeller，这里有一个优化的完全编译解决方案（operator= 委托给另一个不检查自分配的命名成员名称，并且是递归实现的）：copy_data
#include <iostream>
template <typename... Ts> class foo;
template <typename Last>
class foo<Last> {
Last last;
public:
foo (Last r) : last(r) { }
foo() = default;
foo (const foo& other) : last(other.last) { }
foo& operator= (const foo& other) {
if (&other == this)
return *this;
last = other.last;
return *this;
}
protected:
void copy_data (const foo& other) {
last = other.last;
}
};
template <typename First, typename... Rest>
class foo<First, Rest...> : public foo<Rest...> {
First first;
public:
foo (First f, Rest... rest) : foo<Rest...>(rest...), first(f) { }
foo() = default;
foo (const foo& other) : foo<Rest...>(other), first(other.first) { std::cout << "[Copy constructor called]\n"; }
foo& operator= (const foo& other) { // Copy assignment operator
if (&other == this)
return *this;
first = other.first;
// foo<Rest...>::operator= (other);
foo<Rest...>::copy_data(other);
std::cout << "[Assignment operator called]\n";
return *this;
}
protected:
void copy_data (const foo& other) {
first = other.first;
foo<Rest...>::copy_data(other);
}
};
int main() {
const foo<int, char, bool> a(4, 'c', true);
foo<int, char, bool> b = a; // Copy constructor works fine.
foo<int, char, bool> c;
c = b; // Copy assignment operator works fine (and optimized).
}
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