enable_if with copy/move assignment operator

Question

我有一个类,我只想在类的类型参数不是分别复制/移动构造时启用复制/移动赋值操作符.所以我试试这个:

#include <type_traits>

template<typename T>
struct Foobar {

    Foobar(T value) : x(value) {}
    Foobar(const Foobar &other) : x(other.x) {}
    Foobar(Foobar &&other) : x(std::move(other.x)) {}

    template<bool Condition = std::is_nothrow_copy_constructible<T>::value,
             typename = typename std::enable_if<Condition>::type>
    Foobar &operator=(const Foobar &rhs) {
        x = rhs.x;
        return *this;
    }

    template<bool Condition = std::is_nothrow_move_constructible<T>::value,
             typename = typename std::enable_if<Condition>::type>
    Foobar &operator=(Foobar &&rhs) {
        x = std::move(rhs.x);
        return *this;
    }

    T x;
};

int main() {
    Foobar<int> foo(10);
    Foobar<int> bar(20);

    foo = bar;
    foo.operator=(bar);

    return 0;
}

现在,Clang给了我以下错误:

enable_if_test.cpp:31:9: error: object of type 'Foobar<int>' cannot be assigned because its copy assignment operator is implicitly
      deleted
    foo = bar;
        ^
enable_if_test.cpp:8:5: note: copy assignment operator is implicitly deleted because 'Foobar<int>' has a user-declared move
      constructor
    Foobar(Foobar &&other) : x(std::move(other.x)) {}
    ^
enable_if_test.cpp:32:9: error: call to deleted member function 'operator='
    foo.operator=(bar);
    ~~~~^~~~~~~~~
enable_if_test.cpp:4:8: note: candidate function (the implicit copy assignment operator) has been implicitly deleted
struct Foobar {
       ^
enable_if_test.cpp:12:13: note: candidate function [with Condition = true, $1 = void]
    Foobar &operator=(const Foobar &rhs) {
            ^
enable_if_test.cpp:19:13: note: candidate function [with Condition = true, $1 = void] not viable: no known conversion from
      'Foobar<int>' to 'Foobar<int> &&' for 1st argument
    Foobar &operator=(Foobar &&rhs) {
            ^
2 errors generated.

现在,我包含了对赋值运算符的显式调用,以展示错误的奇怪性.它只是说我的模板是一个候选函数,并没有说明为什么没有选择它.

我的猜测是,由于我定义了一个移动构造函数,编译器隐式删除了复制赋值运算符.由于它被隐式删除,它甚至不想实例化任何模板.然而,为什么它表现得像我不知道.

我怎样才能实现这种行为？

(注意:实际代码有合理的理由来定义每个成员,我知道这里没用.)

Answer 1

执行此操作的最佳和唯一方法是通过零规则 - 使用编译器提供的赋值运算符和构造函数,它们复制或移动每个成员.如果成员T x无法复制(移动)分配,则您的类的复制(移动)赋值运算符将默认删除.

SFINAE不能用于禁用复制和/或移动赋值运算符的原因是SFINAE需要模板上下文,但复制和移动赋值运算符是非模板成员函数.

用户声明的拷贝赋值operator X::operator=是一个非静态的非模板的成员函数class X与类型的只有一个参数X,X&,const X&,volatile X&或const volatile X&.

由于您的模板版本不算作用户声明的复制(移动)赋值运算符,因此它们不会禁止生成默认值,并且因为非模板是首选,所以默认模板优先于模板定义(当参数是a const Foobar&,否则模板是更好的匹配,但禁用模板仍然不会禁用自动生成的模板).

如果除了调用成员的复制(移动)赋值运算符之外还需要一些特殊逻辑,请在子对象中实现它(基础或成员都是可行的).

您可以通过选择用作基类的类模板的特化来实现目标,并在继承时传递适当的类型特征:

template<bool allow_copy_assign, bool allow_move_assign>
struct AssignmentEnabler;

template<typename T>
struct Foobar : AssignmentEnabler<std::is_nothrow_copy_constructible<T>::value, 
                                  std::is_nothrow_move_constructible<T>::value>
{
};

当且仅当所选AssignmentEnabler基类具有时,派生类型将使用零规则来默认具有复制和移动赋值.您必须专注AssignmentEnabler于四种组合中的每一种(既不复制也不移动,不移动复制,无复制移动,两者都有).

完整转换您问题中的代码:

#include <type_traits>

template<bool enable>
struct CopyAssignmentEnabler {};

template<>
struct CopyAssignmentEnabler<false>
{
    CopyAssignmentEnabler() = default;
    CopyAssignmentEnabler(const CopyAssignmentEnabler&) = default;
    CopyAssignmentEnabler(CopyAssignmentEnabler&&) = default;
    CopyAssignmentEnabler& operator=(const CopyAssignmentEnabler&) = delete;
    CopyAssignmentEnabler& operator=(CopyAssignmentEnabler&&) = default;
};

template<bool enable>
struct MoveAssignmentEnabler {};

template<>
struct MoveAssignmentEnabler<false>
{
    MoveAssignmentEnabler() = default;
    MoveAssignmentEnabler(const MoveAssignmentEnabler&) = default;
    MoveAssignmentEnabler(MoveAssignmentEnabler&&) = default;
    MoveAssignmentEnabler& operator=(const MoveAssignmentEnabler&) = default;
    MoveAssignmentEnabler& operator=(MoveAssignmentEnabler&&) = delete;
};

template<typename T>
struct Foobar : CopyAssignmentEnabler<std::is_nothrow_copy_constructible<T>::value>,
                MoveAssignmentEnabler<std::is_nothrow_move_constructible<T>::value>
{
    Foobar(T value) : x(value) {}

    T x;
};

int main() {
    Foobar<int> foo(10);
    Foobar<int> bar(20);

    foo = bar;
    foo.operator=(bar);

    return 0;
}