std::map const troubles

[Grey Plastic]

I do not understand why this code fails to compile (under gcc):

#include <map>
using namespace std;

class Foo {
map<Foo*,int> myMap;
public:
int lookup(const Foo& f) const { myMap.find(&f); }
};

The error message states that I am discarding qualifiers by passing a
const map<...> to the map::find function. But there is a const
version of the map::find function. Why isn't the compiler using that
version?

If I replace "map<Foo*,i nt>" with "map<const Foo*,int>" or if I change
"lookup(con st Foo& f)" with "lookup(Foo & f)" suddenly it works. Why?
I am befuddled.

How do I get this to work, without changing the signatures?

I do not understand why this code fails to compile (under gcc):

#include <map>
using namespace std;

class Foo {
map<Foo*,int> myMap;
public:
int lookup(const Foo& f) const { myMap.find(&f); }
};

lookup it supposed to return an int and I dont see it returning one.

Raj

[John Harrison]

"Grey Plastic" <gr*********@ho tmail.com> wrote in message
news:1d******** *************** ***@posting.goo gle.com...

I do not understand why this code fails to compile (under gcc):

#include <map>
using namespace std;

class Foo {
map<Foo*,int> myMap;
public:
int lookup(const Foo& f) const { myMap.find(&f); }
I assume you mean something like this

int lookup(const Foo& f) const { return myMap.find(&f)->second; }
};

The error message states that I am discarding qualifiers by passing a
const map<...> to the map::find function. But there is a const
version of the map::find function. Why isn't the compiler using that
version?
It is using the const version, but the const version takes a Foo* parameter
and you are giving it a const Foo* parameter.

If I replace "map<Foo*,i nt>" with "map<const Foo*,int>" or if I change
"lookup(con st Foo& f)" with "lookup(Foo & f)" suddenly it works. Why?
See above.
I am befuddled.

How do I get this to work, without changing the signatures?

int lookup(Foo f) const { return myMap.find(&f)->second; }

That works, at the cost of copying a Foo object.

john

[Rob Williscroft]

Grey Plastic wrote in
news:1d******** *************** ***@posting.goo gle.com in comp.lang.c++:

I do not understand why this code fails to compile (under gcc):

#include <map>
using namespace std;

class Foo {
map<Foo*,int> myMap;
public:
int lookup(const Foo& f) const { myMap.find(&f); }
};

The error message states that I am discarding qualifiers by passing a
const map<...> to the map::find function. But there is a const
version of the map::find function. Why isn't the compiler using that
version?

If I replace "map<Foo*,i nt>" with "map<const Foo*,int>" or if I change
"lookup(con st Foo& f)" with "lookup(Foo & f)" suddenly it works. Why?
I am befuddled.

How do I get this to work, without changing the signatures?

The key type of the map is `Foo * const` your lookup() function passes
in a `Foo const *` this cast *requires* const_cast<>, as even though
you are adding a const qualifier, you are also removing one.

IOW:

Foo const * -> Foo const * const, is ok but you are doing
Foo const * -> Foo * -> Foo * const.

The good news is that since std::map doesn't dereference the
'key' you can safely use const_cast<>().
int lookup( Foo const &f )
{
myMap.find( const_cast< Foo * >( &f ) );
}

HTH.

Rob.
--
http://www.victim-prime.dsl.pipex.com/

[John Harrison]

> >

How do I get this to work, without changing the signatures?

int lookup(Foo f) const { return myMap.find(&f)->second; }

That works, at the cost of copying a Foo object.

It's rubbish. It compiles but it won't work because the address of the
object passed to find will be different from the address of the object used
to call lookup. Apologies.

john