SFINAE离开了副本构造函数 [英] SFINAE away a copy constructor
问题描述
在某些情况下,我想SFINAE删除类模板的复制构造函数和复制赋值运算符。但是,如果这样做,则会生成默认的复制构造函数和默认的赋值运算符。 SFINAE是基于我作为类模板参数传递的标签完成的。问题是SFINAE仅适用于模板,而复制构造函数/赋值运算符不能用作模板。是否存在解决方法?
Under certain conditions, I'd like to SFINAE away the copy constructor and copy assignment operator of a class template. But if I do so, a default copy constructor and a default assignment operator are generated. The SFINAE is done based on tags I pass as class template parameters. The problem is, that SFINAE only works on templates and a copy constructor/assignment operator can't be a template. Does there exist a workaround?
推荐答案
此解决方案使用有条件不可复制的基类(通过显式标记复制构造函数和
This solution uses a base class that is conditionally not copyable (by explicitely marking the copy constructor and copy assignment operator as deleted).
template <bool>
struct NoCopy;
template <>
struct NoCopy<true>
{
// C++11 and later: marking as deleted. Pre-C++11, make the copy stuff private.
NoCopy(const NoCopy&) = delete;
NoCopy& operator=(const NoCopy&) = delete;
protected:
~NoCopy() = default; // prevent delete from pointer-to-parent
};
template <>
struct NoCopy<false>
{
// Copies allowed in this case
protected:
~NoCopy() = default; // prevent delete from pointer-to-parent
};
示例用法:
template <typename Number>
struct Foo : NoCopy<std::is_integral<Number>::value>
{
Foo() : NoCopy<std::is_integral<Number>::value>{}
{
}
};
int main()
{
Foo<double> a;
auto b = a; // compiles fine
Foo<int> f;
auto g = f; // fails!
}
注意: NoCopy $ c的析构函数$ c>被声明为
受保护的
,以避免虚拟继承(感谢提示,@ Yakk)。
Note: the destructor of NoCopy
is declared protected
to avoid virtual inheritance (Thanks for the hint, @Yakk).
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