Re: How are objects inserted into a set?

From:
 James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Fri, 08 Jun 2007 15:42:40 -0700
Message-ID:
<1181342560.099722.7450@n4g2000hsb.googlegroups.com>
On Jun 7, 9:03 pm, "Victor Bazarov" <v.Abaza...@comAcast.net> wrote:

desktop wrote:

Hmm... Yes. Do you understand std::map? std::set is very similar to
std::map. Essentially 'std::set<T>' is just like 'std::map<const T,T>'.


More like std::map<T, void>, no? T is the key, and there is no
more.

Inserting into a vector works
fine:
class test {
public:
int getpp(){return pp;}
void setpp(int i){pp = i;}
private:
int pp;
};

int main() {
std::vector<test> hh;
test t1;
hh.push_back(t1); // Works fine

std::set<test> my_set;
const test& tref = t1; // see *
my_set.insert(tref); // fails with error: no match for
                   ?operator<? in ?__x < __y?
}

Can I only insert into std::set if my class 'test' define '<' and
properly some of the other operators?


"properly"? Yes, to use the default sorting mechanism your class
needs to have operator< defined for it.


Strictly speaking, std::less<T> must be defined. Normally, this
is done by defining operator<, and allowing the generic
implementation std::less to do its job, but technically, you can
explicitly instantiate std::less directly for your type. (Note
that operator< is only defined on pointers if they point into
the same object, but you can have a set of pointers anyway,
because an implementation is required to make std::less work for
pointer types.)

You can make it a member or you can make it a stand-alone
function.

You don't have to have operator< defined if you use custom sorting
functor in your set.


Which is probably the more usual solution, unless the type has a
real unique ordering.

I still don't see how insert gets the key from 'test' so it can put it
the right place in the tree.


What book are you reading that doesn't explain how sorting of
objects works?


Note that it's very important that the ordering function meet
the specified requirements. In particular, it must be
transitive, and for every a and b, if a<b, then ! b<a, and vice
versa. (Both can, however, be false, in which case the objects
are considered equal.)

This seems trivially obvious, but people are constantly getting
it wrong.

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