转发模板优先于过载 [英] Forwarding template taking precedence over overload

问题描述

我认为，如果参数也匹配，那么非模板将始终优先于模板.

I thought that a non-template would always take precedence over a template, if the arguments match up just as well.

但是:

template <typename... Args>
void Trace(Args&&... args)
{
    throw "what the frak";
}

void Trace(const int&)
{}

int main()
{
    Trace(42);
}

这会抛出，除非我将非模板设为Trace(int)或Trace(int&&) ，即不使用const引用.

This throws unless I make the non-template Trace(int) or Trace(int&&), i.e. not taking a const ref.

这很烦人，因为我想为特定参数类型提供不同的实现，而模板的实际实现是没有意义的(实际上是不会编译的).

It's kind of annoying because I want to provide a different implementation for specific argument types where the real implementation of the template does not make sense (and, in fact, would not compile).

我可以通过将第二个函数专门化来解决它，但前提是我通过使用右值引用来匹配参数类型:

I can fix it by making the second function a specialisation, but only if I match argument types by taking an rvalue reference:

template <>
void Trace(int&&)
{}

而且，也许我不想这样做！也许我有时想用const来称呼它，而宁愿不专攻两次.

And, well, maybe I didn't want to do that! Maybe I want to call it with a const thing sometimes and would rather not specialise twice.

在此示例中，仅提供非模板重载以这种方式工作是否正确并且符合标准?这是我第一次注意到这种行为(并非一定意味着任何事情！).

Is it correct and standard-abiding that simply providing a non-template overload works this way, in this example? It's the first time I've noticed this behaviour (not that that necessarily means anything!).

推荐答案

42是类型int的右值，因此它与int&&的绑定更紧密，而不是与const int&的绑定.这就是为什么它正在调用您的模板Trace函数.

42 is an rvalue of type int, therefore it binds more closely to int&& rather than const int&. This is why it is calling your template Trace function.

如果您要致电

const int i{};
Trace(i);

然后将调用您的Trace(const int&)重载.

可能的解决方法:

• 添加一个调用Trace(const int&)的Trace(int&&)重载.您可能还需要Trace(int&);

• Add a Trace(int&&) overload that invokes Trace(const int&). You might also need a Trace(int&);

在模板化的Trace上使用SFINAE防止在第一个参数为int时实例化；

Use SFINAE on the templated Trace to prevent instantiation when the first argument is an int;

template <typename Arg, typename... Args>
auto Trace(Arg&& arg, Args&&... args)
    -> std::enable_if_t<!std::is_same_v<std::decay_t<Arg>, int>>
{
    throw "what the frak";
}   

更改模板化的Trace以改为使用const Args&....

Change the templated Trace to take const Args&... instead.

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