显式范围-v3十进制类型评估为无效? [英] Explicit range-v3 decltype evaluates to void?

问题描述

我正在尝试获取范围的显式类型(将来我可能希望将其存储为类中的字段).但是，由于某种原因，它的值为void?

I am trying to get an explicit type of a range (I may want to store it as a field in a class in the future). However, for some reason, it evaluates to void?

#include <iostream>
#include <set>
#include <range/v3/view/transform.hpp>

class Alpha {
public:
  int x;
};

class Beta : public Alpha {

};

class Foo {
public:
  std::set<Alpha*> s;

  using RangeReturn = decltype(std::declval<std::set<Alpha*>>() | ranges::v3::view::transform(std::function<Beta*(Alpha*)>()));
  RangeReturn r();
};

Foo::RangeReturn Foo::r() {
  return s | ranges::v3::view::transform([](Alpha* a) { return static_cast<Beta*>(a); });
}

int main() {
}

使用g ++ -std = c ++ 17进行编译时，它会给出

When compiling with g++ -std=c++17, it gives

main.cpp:24:88: error: return-statement with a value, in function returning 'void' [-fpermissive]

(g ++版本g ++(Ubuntu 7.3.0-27ubuntu1〜18.04)7.3.0)

(g++ version g++ (Ubuntu 7.3.0-27ubuntu1~18.04) 7.3.0)

我在Visual Studio 2017 15.9版中收到类似错误

I get an error of similar kind on Visual Studio 2017, v. 15.9

这个问题是我其他问题的延续:

This question is a continuation of my other question: How to store a range as a field in a class? but is more specific and I believe it deserves to be separate.

推荐答案

您的代码不起作用，原因是:

Your code doesn't work because:

• range/v3视图从右值禁用视图，因为这将导致悬空引用.因此，在您的declval()中，您还应该使用左值:

• range/v3 view disables view from rvalue, because that will cause dangling reference. Thus in your declval(), you should also use an lvalue:

std::declval<std::set<Alpha*>&>()
//                           ^ here should be lvalue

视图转换信息被编码在template参数内.因此，如果使用view::transform(std::function<Beta*(Alpha*)>())表示类型，则表达式应具有完全相同的类型. Lambda不好.

view transformation information is encoded inside the template parameter. So if you use view::transform(std::function<Beta*(Alpha*)>()) to represent the type, your expression should have exactly the same type. A lambda is not ok.

有效版本为:

class Foo {
public:
  std::set<Alpha*> s;

  using RangeReturn = decltype(std::declval<std::set<Alpha*>&>() | ranges::v3::view::transform(std::function<Beta*(Alpha*)>()));
  RangeReturn r();
};

Foo::RangeReturn Foo::r() {
  return s | ranges::v3::view::transform(std::function<Beta*(Alpha*)>{
          [](Alpha* a) { return static_cast<Beta*>(a); }
          });
}

但是实际上，以这种方式存储视图不是一个好主意.

But actually, it's not a good idea to store a view this way.

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