类方法的返回类型演绎？ C ++ 1Y [英] Return type deduction for class methods? C++1y

问题描述

我问，因为我有点含蓄地认为它会工作，但在gcc 4.8.1中，我得到一个内部编译器错误（in gen_type_die_with_usage）。我第一次得到这样一个神秘的错误，所以我有点怀疑;我知道自那以后他们已经改变了规范。

为了清楚起见，这适用于我：

  auto foo（）{return 5;} 
  

：

  class Bar {
 auto baz（）{return 5;} 
} 
  

标准草案中允许这样做吗？

解决方案

是的标准应该允许它根据文件 n3582 。这是一篇来自论文的例子。
$ b

允许使用自动返回类型的非定义函数声明不是必须的，但它对于更喜欢定义成员函数的编码样式在类之外：

  struct A {
 auto f（）; //前置声明
}; 
 auto A :: f（）{return 42; } 
  

如果我们在这种情况下允许它，它应该是有效的在其他情况下也是如此。允许它也是更正交的选择;总的来说，我相信如果将两个功能结合起来就可以发挥作用。根据@bamboon的评论，返回类型扣除仅在gcc 4.9支持。

所以这可以解释为什么你没有它。

Is return type deduction allowed for member functions in c++14, or only for free functions?

I ask because I sort of implicitly assumed it would work, but in gcc 4.8.1 I get an internal compiler error("in gen_type_die_with_usage"). First time I have ever gotten such a cryptic error like that, so I am a bit skeptical; and I know they have changed the spec since then.

For clarity this works for me:

auto foo() {return 5;}

but this doesn't:

class Bar{
auto baz() {return 5;}
}

Is this allowed in the draft standard?

解决方案

Yes the standard should allow it according to the paper n3582. Here is an example from the paper.

Allowing non-defining function declarations with auto return type is not strictly necessary, but it is useful for coding styles that prefer to define member functions outside the class:

    struct A {
      auto f(); // forward declaration
    };
    auto A::f() { return 42; }

and if we allow it in that situation, it should be valid in other situations as well. Allowing it is also the more orthogonal choice; in general, I believe that if combining two features can work, it should work.

According to the comment by @bamboon, "Return type deduction is only supported as of gcc 4.9." so that would explain why you don't have it.

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