我不了解std :: result_of和decltype的这些用法 [英] I'm not understanding these uses of std::result_of and decltype

问题描述

在我的Visual Studio项目中，我有以下内容，并且运行良好：

On my Visual Studio project I had the following and it worked fine:

template <typename T>
void function(T type)
{

    std::result_of<T()>::type myVar = 5; // This compiled fine with Visual Studio, 
but with GCC it shows the following errors:
    //error: dependent-name ‘std::result_of<T()>::type’ is parsed as a non-type, 
    //but instantiation yields a type. note: say ‘typename std::result_of<T()>::type’ if a type is meant
}

int main() {

    auto lambda = []() { return float(); };
    function(lambda);
    return 0;
}

我只是想了解一下，是编译器坚持要求我为std作序： ：result_of具有类型名，因为它可能是含糊的，在该std :: result_of中可以返回一个类，然后:: type可以成为该类的成员吗？这就是为什么它坚持要添加类型名称吗？如果是这种情况，那么为什么Visual Studio允许它？还不合规吗？

I just want to understand, is the compiler insisting that I preface the std::result_of with "typename" because it could be ambiguous, in that std::result_of could return a class, and then ::type could be a member of that class? Is that why it's insisting that typename be added? If this is the case then why does Visual Studio allow it? Is it non compliant?

此外，由于我已经阅读到自C ++ 14或C ++ 17起不推荐使用result_of，因此我想尝试并使用更通用的decltype，应该在更多情况下使用。所以我尝试了：

Also, because I've read that result_of is being deprecated as of C++14 or C++17, I wanted to try and use the more generic decltype, which is supposed to work in more circumstances. So I tried:

template <typename T>
void function(T type)
{
decltype(T()) myVar = 5; // Error, use of deleted function‘main()::<lambda()>::<lambda>()’  main.cpp    
}

所以我知道lambda具有已删除的默认构造函数和复制赋值运算符，但是在这种情况下，我确实认为将lambda传递给此模板时函数lambda的副本构造函数被调用，它确实具有。然后，当我执行decltype（T（））时，我假设这将调用其operator（）函数。我不明白为什么它会说一些有关删除功能的信息。

So I'm aware that a lambda has a deleted default constructor and copy assignment operator, but in this case I really think that when passing the lambda to this templated function the lambda's copy constructor is called, which it does have. Then when I do decltype(T()) I assume this would be calling its operator() function. I don't understand why it says something about deleted function.

最后我尝试了：

decltype(std::declval<T()>) myVar = 5;

因为我认为declval可以像在调用时创建假实例一样使用至少那是对我的解释。这也会因错误而失败：

Because I thought that declval can be used as if creating a fake instance of whatever call you make, at least that's how it was explained to me. This also fails with error:

类型为'main（）::&&'
的引用的无效初始化从类型'int'

"invalid initialization of reference of type ‘main()::&&’ from expression of type ‘int"

推荐答案

result_of

首先，GCC编译器在 std :: result_of typename $ c>，因为后者的返回值是一个类。并且必须指示它使用其类型来声明新变量。

First, the GCC compiler requires the keyword typename before std::result_of because the return value of the latter is a class. And you have to instruct it to use its type to declare a new variable.

关于您的评论：

另外，因为我已经阅读到自C ++ 14或C ++ 17起不推荐使用result_of

Also, because I've read that result_of is being deprecated as of C++14 or C++17

从C ++ 17开始不推荐使用

std :: result_of （在这里查看原因），并用新引入的 std :: invoke_result 替换为
，所以您如果您有兼容的编译器，则可以改用它。

std::result_of is deprecated as of C++17 (See here why) and is replaced by the newly introduced std::invoke_result, so you could use it instead if you have a compliant compiler.

decltype

std :: result_of 是按照 decltype 的以下方式声明的：

Since std::result_of is declared in terms of decltype in the following manner:

  template<typename _Signature>
  struct result_of;

  template<typename _Functor, typename... _ArgTypes>
  struct result_of<F(Args...)> {
      typedef decltype( std::declval<F>()(std::declval<Args>()...) ) type;
  };

您可以使用类似的定义：

You could use a similar definition:

decltype( std::declval<T>()() ) myVar = 5;

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