使用并重载基类的模板成员函数? [英] using and overloading a template member function of a base class?

问题描述

以下，结构Y重载了X的成员函数f.这两个重载都是模板函数，但是采用不同的参数(typename和int)进行明确指定:

In the following, struct Y overloads X's member function f. Both overloads are template functions, but take different arguments (typename and int), to be explicitly specified:

struct X
{
    template <typename> static bool f() { return true; }
};

struct Y : public X
{
    using X::f;
    template <int> static bool f() { return false; }
};

int main()
{
    std::cout << Y::f <void>() << " " << Y::f <0>() << std::endl;
}

这将按预期使用gcc打印1 0.但是，clang(3.3)抱怨

This prints 1 0 using gcc, as expected. However, clang (3.3) complains that

[...] error: no matching function for call to 'f'
        std::cout << Y::f <void>() << " " << Y::f <0>() << std::endl;
                     ^~~~~~~~~~~
[...] note: candidate template ignored: invalid explicitly-specified argument
      for 1st template parameter
        template <int> static bool f() { return false; }
                                   ^

即只能看到Y的版本.我已经尝试过

i.e., can only see Y's version. I've tried

using X::template f;

相反，没有成功.非静态(模板)成员函数也会发生同样的情况.那是一个错误吗?

instead, with no success. The same happens for non-static (template) member functions. So is this a bug?

推荐答案

这个难题最近是根据另一个答案向我解释的.

This conundrum was recently explained to me in the light of another answer.

在#clang IRC频道中:

From the #clang IRC channel:

[01:16:23] <zygoloid> Xeo: this is a weird corner of the language where clang conforms but the rule is silly
[01:16:31] <Xeo> ... really? :(
[01:16:45] <zygoloid> Xeo: when deciding whether a using-declaration is hidden, we're not allowed to look at the template-parameter-list (nor the return type, iirc)
[01:17:04] <zygoloid> so the derived class declaration of operator()(T) suppresses the using-declaration
[01:17:19] <Xeo> because it has the same signature / parameter types?
[01:17:40] <zygoloid> rigth

解决方法是不在uses派生版本的类中定义f.而是将其移到辅助帮助程序类中(在这种情况下，这是一个问题，您认为应该赢得哪个定义).

The workaround is to not define f in the class that uses the derived version. Instead, move it into an auxiliary helper class (which, in this case begs the question, which definition you reckon should win).

• 请参阅此处了解我之前遇到的问题: Lambda函数作为基础类

@@ Xeo知道为什么c ++++拒绝这样做: coliru.stacked-crooked.com/a/6a0e6a1cac062216(clang: coliru.stacked-crooked.com/a/5c2d6dd449a92227)

@Xeo any idea why clang++ refuses this: coliru.stacked-crooked.com/a/6a0e6a1cac062216 (clang: coliru.stacked-crooked.com/a/5c2d6dd449a92227)

以及如何使用额外的基类对其进行修复:

And here how to fix it using an extra base-class:

@Xeo还是修复了它，由于某种原因，这种形式不适合clang ++ stackoverflow.com/a/18432618/85371

信用:感谢@Xeo和休息室中的人员挖掘出这个愚蠢的规则"

Credits Thanks to @Xeo and people in the Lounge for unearthing this "silly rule"

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