Question

我有这样一段代码：

template<template<typename> class ContainerType,
         typename ValueType,
         typename ReturnType>
struct value_extractor
{
public:
    static ReturnType extract(const ContainerType<ValueType>&);
};

template<template<typename> class ContainerType,
         typename ValueType>
struct value_extractor<ContainerType, ValueType, std::shared_ptr<ValueType>>
{
    static std::shared_ptr<ValueType> extract(const ContainerType<ValueType>& value)
    {
        return value;
    }
};

实质上是从模板类型中提取值。这段代码使用clang编译良好，但是使用gcc时出现错误消息：

g++ test.cpp -lstdc++ -O2 
In file included from di.hpp:1:0,
                 from test.cpp:2:
holders.hpp: In instantiation of ‘ReturnType di::holder<ContainerType, ValueType>::provide() const [with ReturnType = std::shared_ptr<int>; ContainerType = std::shared_ptr; ValueType = int]’:
di.hpp:35:105:   required from ‘static ReturnType di::holder_selector::convert(const types_map&, ContainerType<ValueType>*) [with ReturnType = std::shared_ptr<int>; ContainerType = std::shared_ptr; ValueType = int; di::types_map = std::unordered_map<void (*)(), std::unique_ptr<di::base_holder> >]’
di.hpp:40:39:   required from ‘T di::injector::provide() const [with T = std::shared_ptr<int>]’
test.cpp:14:63:   required from here
holders.hpp:48:85: error: ‘di::value_extractor<ContainerType, ValueType, std::shared_ptr<_Up> >::extract(const ContainerType<ValueType>&) [with ContainerType = std::shared_ptr; ValueType = int]’ is not a template [-fpermissive]
 alue_extractor<ContainerType, ValueType, ReturnType>::template extract(value_);
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~

如果我使用-fpermissive标志，则代码可以编译甚至可以工作，但显然会发出警告。所以我的问题是：真的是我还是这是gcc的错误，如果是我编写了不一致的代码，那么我应该如何解决呢？预先感谢。

Answer 1

调用提取，愚蠢的错误时，不需要

template。感谢@songyuanyao

为什么gcc告诉我我的模板不是模板？

1 个答案: