c ++模板函数专门化 - 模板参数数错误？ [英] c++ template function specialization - wrong number of template arguments?

问题描述

好吧，所以我只是在学习模板。无论如何，我可能（绝对）做错了，但这里的问题：

Okay, so I'm just learning about templates. Anyways, I'm probably (most definitely) doing something wrong, but here's the problem:

我的第一个模板函数声明如下：

My first template function is declared like this:

template<typename T>
std::ostream& printFormatted(T const& container, std::ostream& os = std::cout) {
    //...
}

然后我应该为地图实现一个专门的例子，所以这是我试图做的：

Then I'm supposed to implement a specialized case for maps, so this is what I tried to do:

template<>
std::ostream& printFormatted<std::map<typename key, typename value>>(std::map<typename key, typename value> const& container, std::ostream& os = std::cout) {
    //...
}

我可能会错误地使用键/值变量，不确定，但无论如何，当试图编译我得到错误消息：

I might be making a mistake with my key/value variables, not sure, but regardless, when trying to compile I get the error message:

error: wrong number of template arguments (1, should be 4)
error: provided for ‘template<class _Key, class _Tp, class _Compare, class _Allocator> class std::__debug::map’

显然有些东西我不明白模板或地图？有人请帮忙吗？

Clearly there's something I don't understand about templates or maps? Someone please help?

推荐答案

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.

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