如何指定返回类的constexpr函数的类型（不使用auto关键字）

问题描述：

基本上在下面我想看看是否可以绕过使用 auto 关键字

Basically in below I want to see if I can get around having to use auto keyword

我们有下面的代码[使用g ++ 4.9.2（Ubuntu 4.9.2-10ubuntu13）& clang version 3.6.0]：

Suppose that we have the following piece of code [works with g++ 4.9.2 (Ubuntu 4.9.2-10ubuntu13) & clang version 3.6.0] :

//g++ -std=c++14 test.cpp
//test.cpp

#include <iostream>
using namespace std;

template<typename T>
constexpr auto create() {
  class test {
  public:
    int i;
    virtual int get(){
      return 123;
    }
  } r;
  return r;
}

auto v = create<int>();

int main(void){
  cout<<v.get()<<endl;
}

如何指定code>而不是在声明/定义点使用 auto 关键字？我尝试了 create< int> :: test v = create< int>（）; ，但这不起作用。

How can I specify the type of v rather than using the auto keyword at its point of declaration/definition? I tried create<int>::test v = create<int>(); but this does not work.

1）这不同于我在从constexpr函数返回一个类需要带g ++的虚拟关键字即使代码是相同的

1)this is different from the question that I was asking at Returning a class from a constexpr function requires virtual keyword with g++ even through the code is the same

2）我不要在函数外定义类。

2)I do not want to define the class outside the function.

答

实际类型是隐藏的，因为它在函数内部是局部的，所以你不能明确地使用它。但您应该能够使用 decltype ，如

The actual type is hidden as it's local inside the function, so you can't explicitly use it. You should however be able to use decltype as in

decltype(create<int>()) v = create<int>();

我没有看到这样的原因，但 / code>工程。

I fail to see a reason to do like this though, when auto works.

如何指定返回类的constexpr函数的类型（不使用auto关键字）

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