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开发者_开发技巧 change key value from a map iterator)?
Your reasoning is correct, value_type::first
would be char const * const
.
There is a common source of confusion in thinking that const T
when T
is a type *
is const type *
, but that is not so. Unlike macros, typedef
s are not text substitution, nor are template arguments. When you do const T
, if T
is a typedef
or template argument, you are adding a const
to the type as a whole.
That is the one reason why I like to write my const
s at the right of the type, as it causes less confusion: T const *
, add an extra const, get T const * const
.
If a
is const char*
, then const a
is indeed const char* const
.
You are correct in your assessment, but you have to be very careful with this approach for two reasons:
- You have to supply a custom comparator predicate that does proper ordering of const char * (e.g., using a variant of strcmp that only returns true if key1 < key2 ). If you do not do this, your strings will be ordered by pointer values, which is almost always not what you want.
- You have to consider the lifetime of the keys in your map and how they will be deallocated when the map is destroyed.
精彩评论