为什么 gcc 告诉我我的模板不是模板?

>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;
}
};

它实质上是从模板类型中提取值。这段代码使用 clang 编译得很好，但使用 gcc 时我收到一个错误，说：

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，代码可以编译甚至工作，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，如果是我编写不合格的代码，那么我应该如何解决它？ 提前谢谢。