std::map and case insensitive search

[bb]

Hi,

void fun(const std::map<std::string,int>& m1) {
// How to make a case insensitive search of this map without
making a copy?
}

cheers.

[red floyd]

bb wrote:

Hi,

void fun(const std::map<std::string,int>& m1) {
// How to make a case insensitive search of this map without
making a copy?
}

cheers.

Don't use std::map<string, int>. Use std::map<string, int, predicate>.

[John Harrison]

bb wrote:

Hi,

void fun(const std::map<std::string,int>& m1) {
// How to make a case insensitive search of this map without
making a copy?
}

cheers.

Essentially what you're asking for is a std::map with two different
keys, (one case sensitive, one case insensitive) which is impossible.

You need to rework your problem. You could try

std::map<case_insensitive_string, int>

or

std::map<std::string, int, case_insensitive_comparator>

or

you could convert all strings to one case before inserting into the map.

If you really need both case sensitive and case insensitive searches on
the same map, you are going to have to come up with a more complex data
structure yourself. You can't expect the standard library to do
everything for you.

john

[Juha Nieminen]

bb wrote:

void fun(const std::map<std::string,int>& m1) {
// How to make a case insensitive search of this map without
making a copy?
}

The map data structure is organized with the default < operator
in this case. It's simply physically impossible, using this map
instance only, to perform a O(log n) search when the comparator
is different because the items simply aren't organized that way.

You can perform a linear search, though (in this case it's not
possible to achieve anything faster). Simply traverse from begin()
to end() and compare keys.

[Juha Nieminen]

John Harrison wrote:

Essentially what you're asking for is a std::map with two different
keys, (one case sensitive, one case insensitive) which is impossible.

Actually he didn't impose any speed limitations on the search.
Nothing stops him from making a linear search through they elements
of the map.