以指针为键的映射的 value_type
据我所知，C++ 将 map::value_type 定义为 pair
如果我使用指针类型作为键类型会发生什么在地图中，即，正如
std::map<const char*,int>::value_type::first_type = const char*
我从上面的定义中所期望的那样，或者
std::map<const char*,int>::value_type::first_type = const char* const
更符合逻辑（因为否则我将被允许从地图迭代器更改键值）？
For what I know, C++ defines map<a,b>::value_type as pair<const a,b>
What will happen if I use a pointer type as key type in map, i.e., is
std::map<const char*,int>::value_type::first_type = const char*
as I would expect from definition above or
std::map<const char*,int>::value_type::first_type = const char* const
as would be more logical (since otherwise i would be allowed to change key value from a map iterator)?
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评论（3）
您的推理是正确的，
value_type::first将是char const * const。有一个常见的混淆来源，即当
T是type *时，认为const T是const type *，但事实并非如此。与宏不同，typedef不是文本替换，模板参数也不是。当您执行const T时，如果T是typedef或模板参数，您将向键入一个整体。这就是为什么我喜欢在类型右侧编写
const的原因之一，因为它会减少混乱：T const *，添加一个额外的 const，得到T const * const。Your reasoning is correct,
value_type::firstwould bechar const * const.There is a common source of confusion in thinking that
const TwhenTis atype *isconst type *, but that is not so. Unlike macros,typedefs are not text substitution, nor are template arguments. When you doconst T, ifTis atypedefor template argument, you are adding aconstto the type as a whole.That is the one reason why I like to write my
consts at the right of the type, as it causes less confusion:T const *, add an extra const, getT const * const.如果
a是const char*，那么const a确实是const char* const。If
aisconst char*, thenconst ais indeedconst char* const.您的评估是正确的，但您必须非常小心这种方法，原因有两个：
You are correct in your assessment, but you have to be very careful with this approach for two reasons: