如何从重载解析中删除一个函数？ [英] How do I remove a function from overload resolution?

问题描述

在使用赋值运算符时，我希望将我的类的实例转换为 int 。所以我写了这段代码：

I want to have an instance of my class convert to an int when using the assignment operator. So I wrote this code:

struct X
{
    X() = default;
    X& operator=(int) { std::cout << "operator=(int)\n"; return *this; }
    operator int() { return 0; }
};

int main()
{
    X a, b;
    a = b;
}

但是它没有被调用。这是因为它正在调用隐式副本分配运算符，该运算符与参数完全匹配。我希望我的代码先在 b 上调用转换运算符，然后将 int 返回值绑定到 operator =（）。

But it does not get called. This is because it's calling the implicit copy-assignment operator which is an exact match for the argument. I want my code to call the conversion operator on b first and then have the int return value bind to operator=().

是否有语法告诉编译器不考虑此功能？换句话说，如何从重载解析中删除函数？

Is there a syntax to tell the compiler "do not consider this function"? In other words, how do I remove a function from overload resolution?

我尝试在复制分配运算符上使用模板，因此可以执行SFINAE，但我想只是创建了另一个函数，因此非模板函数总是会更好地匹配。

I've tried using templates on the copy-assignment operator so I could do SFINAE but I guess that just creates another function so the non-template one is always going to be a better match.

推荐答案

隐式创建的赋值运算符保持不变最佳版本，即使您 =删除也是如此，例如，您将得到一个错误，而不是选择了另一个版本。我建议仅转发适当的逻辑，而不必费心尝试删除该功能：

The implicitly created assignment operator stays the best version even if you = delete it, i.e., you'd get an error instead of another version being chosen. I would recommend simply forwarding the logic as appropriate and not bother trying to remove the function:

X& X::operator= (X const& other) {
    return (*this) = static_cast<int>(other);
}

诚然，这潜在地允许额外的隐式转换序列，否则可能不允许：

Admittedly, this potentially allows an extra implicit conversion sequence which may not otherwise be allowed:

T -> X -> int

要应对这种可能性，您可以添加另一个赋值运算符：

To counter this possibility, you'd add another assignment operator:

template <typename T>
X& X::operator= (T&& other) {
    int arg = other;
    return (*this) = arg;
}

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