Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
854 views
in Technique[技术] by (71.8m points)

c++ - How to specify type of a constexpr function returning a class (without resorting to auto keyword)

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

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;
}

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.

p.s.

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

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

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>();

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


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...