假设您使用键 c $ c> value 意味着是placeholers，不能使用关键字 typename 来声明模板参数。也就是说， Foo< typename T> 总是无效的，但不能被 Foo< typename T :: bar& code>这是完全不同的。专用化语法如下：

Assuming your uses of key and value are meant to be placeholers, you cannot declare template parameters inline with the keyword typename. That is to say, Foo<typename T> is always invalid -- but not to be mistaken with Foo<typename T::bar> which is different altogether. The syntax for specialization looks like:

// Declare template parameters up front
template<typename Key, typename Value>
std::ostream&
printFormatted<std::map<Key, Value> >(std::map<Key, Value> const& container, std::ostream& os = std::cout);

但这不会工作，因为它是一个部分不允许用于函数模板。使用重载：

but that wouldn't work because it's a partial specialization and those are not allowed for function templates. Use overloading instead:

template<typename Key, typename Value>
std::ostream&
printFormatted(std::map<Key, Value> const& container, std::ostream& os = std::cout);

此重载将优先于更一般的模板。

This overload will be preferred over the more general template.