Re: Modifying the value of key of a map.

From:
homsan toft <nowhere@specific.org>
Newsgroups:
comp.lang.c++
Date:
Tue, 30 May 2006 13:38:26 +0200
Message-ID:
<e5hari$fcc$2@oden.abc.se>
Milind wrote:

Hi,

I have a map of the type:

std::map<A, B> mymap_A;
std::map<X, A> mymap_X;

where A, B and X are user defined types (i.e. my own clases)

At the beging of the program i do a insert on the maps. Later, at some
point i come across a situation where the value of some member of A
changes. e.g:

class A {
int a_member;
};

when i did an insert a_member was 2; now it is supposed to be 4.

Problem:

my first, and absolutely naive approach was:

std::map<A,B>::iterator i = mymap.find(valueofkey);
i->first = 4;

But this is not allowed as i->first is read only.

Now, the possible solution for this would be to either create a new
pair (and necessarily copy everything in the value part... which is
going to be very costly) and delete old and insert new pair.

Am I missing something too obvious??


It's hard to tell without looking into your intentions, and perhaps at classes B and X.
Maybe you mixed up the Key and Value parts, or you want something other than a map<A,B>, or...

The basic idea of a map<Key,Value> is that you do (fast) lookup by key,
but you can edit the data freely.
Like
     mymap[player.id()].lives--;
or
     mymapiter i = mymap.find(player.id()); // mymapiter is supposed to be a (badly named) typedef
     if (i != mymap.end()) // if you use find, you must test, otherwise you crash on dereference
         i->lives--;

player.id() is the key here. The data is a struct with an int-like member 'lives'.
If you use any other key, you should be looking at other data, ie the data of another player.
And obviously, the id shouldn't change (if it does "id" is probably a bad name for it).
This is what std::maps are supposed to be good for.

If you want to look up by data, eg to find all players with zero lives,
you need something different (perhaps a bidirectional map - is this why you
have both map<A,B> and map<X,A> ? - no way for us to know...)

HTH
    homsan

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