用户定义的类作为模板参数 [英] User Defined Class as a Template Parameter
问题描述
我正在重新实现 std :: map
.我需要确保任何数据类型(基本或用户定义的)键都可以使用.我将Map类声明为模板,该模板具有用于键和值的两个参数.我的问题是,是否需要使用字符串作为键类型,如何重载<和>仅适用于字符串类型键的运算符?在模板专门化中,我们必须按照我的理解将整个类专门化为需要的类型.
有什么办法可以更好地做到这一点?如果我添加一个单独的Key类并将其用作Key的模板类型怎么办?
I' m re-implementing std::map
. I need to make sure that any data type (basic or user defined) key will work with it. I declared the Map class as a template which has two parameters for the key and the value. My question is if I need to use a string as the key type, how can I overload the < and > operators for string type keys only?? In template specialization we have to specialize the whole class with the type we need as I understand it.
Is there any way I can do this in a better way?? What if I add a separate Key class and use it as the template type for Key?
推荐答案
您应该像普通的 std :: map
一样,将比较列为一种类型.也就是说,具有实用程序类 less_compare
:
You should factor out the comparison as a type, like the normal std::map
does. That is, have a utility class less_compare
:
template <typename T>
struct less_compare
{
bool operator()(const T& pLhs, const T& pRhs) const
{
return pLhs < pRhs;
}
};
然后:
template <typename Key, typename Value, typename Compare = less_compare<Key> >
class map
{
// ...
private:
Compare mCompare;
};
要比较两个值,请执行以下操作: if(mCompare(someThing,someOtherThing))
,对于 someThing
来说,它是小于" someOtherThing
.注意,该分解也允许用户定义比较(这就是为什么引用小于"的原因).这就是基于策略的设计.
And to compare two values, do: if (mCompare(someThing, someOtherThing))
, which will be true with someThing
is "less than" someOtherThing
. Note this factoring also allows user-defined comparisons (which is why "less than" is quoted). This is known as policy-based design.
现在您可以仅将C字符串的 less_compare
类专门化.(并且还提供 greater_compare
和亲属.)
And now you can specialize just the less_compare
class for C-strings. (And also provide greater_compare
and kin.)
请记住,除非这是为了学习,否则您不应该实现自己的地图.还要注意, std :: string
的 operator<
已经过载.
Do keep in mind, unless this is for learning, you should not be implementing your own map. Also note that std::string
has operator<
overloaded already.
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