根据数字模板参数有条件地编译转换运算符 [英] Conditionally compile a conversion operator based on numeric template parameter

问题描述

我有一个模板< bool P> Foo 类，其中包含大量代码。我希望能够将 Foo< true> 转换为 Foo< false> ，即运算符Foo< false>（）方法。但是编译器不喜欢Foo现有的这种方法，它只喜欢 Foo< true> 的方法，并警告如何不调用运算符用于隐式或显式转换（GCC 5.4.x）

I have a template <bool P> class Foo with lots of code in it. I want to be able to convert Foo<true>'s into Foo<false>'s, i.e. have an operator Foo<false>() method. But the compiler doesn't like such a method existing for Foo, it only likes it for Foo<true>, and gives a warning about how the "operator will not be called for implicit or explicit conversions" (GCC 5.4.x)

似乎我不能为此使用SFINAE： std :: enable_if 适用于类型；我尝试了一个值变量（真实情况下的值是 value 而不是 type 成员）没有

It doesn't seem like I can use SFINAE for this: std::enable_if works with types; and a value-variant I tried out (where the true case has a value rather than a type member) didn't help either.

如何才能仅针对 Foo< false> （除了

到目前为止，我最好的尝试是：尝试将 Foo< false> 不同并复制我的所有代码）？

My best attempt so far has been:

template <bool P> class Foo {
    // etc. etc.
    template <bool OtherValue>
    operator Foo<OtherValue>()
    {
        static_assert(OtherValue, "You should not be using this conversion!");
        // conversion code here
        return Foo<false>(args,go,here);
    }
}

推荐答案

如何才能仅针对 Foo< false> 编译此运算符（而不是专门针对 Foo< false> 不同并复制我所有的代码）？

How can I get this operator to only be compiled for Foo<false> (other than specializing Foo<false> different and duplicating all of my code)?

template <bool P> 
struct Foo 
{
    template <bool P2 = P, typename = std::enable_if_t<P2 == true>>
    operator Foo<false>()
    {
        return {};
    }
};

使用 P2 = P 会延迟对 enable_if_t 到实际使用转换运算符的时间（而不是类实例化）。

Using P2 = P delays the evaluation of the enable_if_t to when the conversion operator is actually being used (instead of class instantiation).

如果我们尝试简单地写：

If we tried to simply write:

template <typename = std::enable_if_t<P == true>>
operator Foo<false>()
{
    return {};
}

std :: enable_if_t< P == true> 将在 Foo< P> 的实例化期间进行评估，因为在实例化成员函数时不会发生替换。实例化 Foo 时发生替换-因此SFINAE无法发生（因为尚未设置任何重载解决方案）。

std::enable_if_t<P == true> would be evaluated during Foo<P>'s instantiation as there is no substitution occurring when the member functions are instantiated. The substitution is happening when Foo is instantiated - therefore SFINAE cannot take place (as there isn't any overload resolution set yet).

通过添加 bool P2 = P 默认参数，我们将替换延迟到转换运算符的实例化上。

By adding a bool P2 = P default parameter, we delay substitution to the instantiation of the conversion operator. This happens during overload resolution, so SFINAE can take place.

此答案比我能更好地解释：https://stackoverflow.com/a/13401982/598696

This answer explains it better than I can: https://stackoverflow.com/a/13401982/598696

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