Template C++: How to access iterator value for both std::map and std::set?

Question

I have a specific search function. Because it's used on both std::set and std::map, it was duplicated in our code (not using template).

I have to maintain those two functions, and I'd like to move them to a single function using a template (and then have only one search preocedure to maintain).

I can't find how to convert an iterator to the container's value_type. For std::set, you just need to dereference the iterator (*iter), but for std::map, you need to access the second item of the iterator (which is a pair) iter->second.

Here is an isolated example:

template <class Container, class Object> bool MyFindFunction( const Container& container, Object& found )
{
    Container::const_iterator iter = container.begin();
    // do my special search procedure here
    // note that this code also needs to access the iterator's value...

    if ( iter != container.end() )
    {
        found = *iter; // this works for set, but not for map
        found = iter->second; // this works for map, but not for set
        // HOW TO MAKE IT WORK FOR BOTH??
        return true;
    }
    else
    {
        return false;
    }
}

int main ()
{
    std::set<double> mySet;
    std::map<int,double> myMap;
    double found = 0;

    MyFindFunction( mySet, found );
    MyFindFunction( myMap, found );
}

Note that the special search procedure also needs to access the value_type (or mapped_type for maps), so, moving this procedure to a template function and having a MyFindFunctionInMap and MyFindFunctionInSet function handling iterator to value conversion after calling the search procedure function won't help.

PS: Sorry, I'm using C++98..

Just create an access function using the templated iterator and implement it for map/set iterator. So you dont have the complete function implementedtwice, just the problematic access — bazz-dee
– bazz-dee, Commented Jun 16, 2015 at 8:34
Pedantic point: the value_type of a std::map<A, B> is std::pair<const A, B>, which is exactly what dereferencing an iterator gives you. The type you're looking for is mapped_type. — Sebastian Redl
– Sebastian Redl, Commented Jun 16, 2015 at 9:52
Yes, I'm looking for mapped_type, which is not defined for sets...this one only has value_type.... — jpo38
– jpo38, Commented Jun 16, 2015 at 9:54

songyuanyao · Accepted Answer · 2015-06-16 10:27:52Z

3

You can use template function overload to distinguish these cases:

template <typename V>
inline V get_value(const V& v) { return v; }
template <typename K, typename V>
inline V get_value(const std::pair<K, V>& p) { return p.second; }

and then

found = get_value(*iter);

LIVE DEMO

edited Jun 16, 2015 at 10:27

answered Jun 16, 2015 at 8:41

songyuanyao

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3 Comments

jpo38 Over a year ago

I like that, but it does not compile (I'm using VS2010). Apparently the map version function is not used by the compiler. Would there by a way to pass the get_value function to be used as a parameter?

songyuanyao Over a year ago

@jpo38 My fault. I've updated my answer, pls check again.

jpo38 Over a year ago

It works great and is really a nice and easy way to fix the problem. Thanks!

Ami Tavory · Accepted Answer · 2015-06-16 09:18:18Z

1

You could use a boost::transform_iterator to build a set-like iterator from the map-like iterator.

Create an internal template function parameterized by a set-like iterator and only taking iterators, not the container, as arguments. Dispatch to this function either the original iterator or the transformed iterator.

Edit If you cannot use boost, build some functor to access your stuff:

#include <map>
#include <set>
#include <iostream>


using namespace std;


template<class Key, class Value>
struct access_key
{
    template<class Ref>
    const Key &operator()(const Ref &v) const
    {
        return v.first;
    }
};

template<class Key>
struct access_key<Key, Key>
{
    template<class Ref>
    const Key &operator()(const Ref &v) const
    {
        return v;
    }
};


template<class Container>
void fn(Container t)
{
    access_key<typename Container::key_type, typename Container::value_type> a;
    cout << a(*t.begin()) << endl;
}


int main()
{
    set<int> s;
    s.insert(1);

    map<int, int> m;
    m[1] = 1;

    fn(s);
    fn(m);

    return 0;
}

This accessor relies on the key type and value type being distinct/same for map/set.

edited Jun 16, 2015 at 9:18

answered Jun 16, 2015 at 8:28

Ami Tavory

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5 Comments

Ami Tavory Over a year ago

@jpo38 See update. In the future, you might want to state restrictions in your question, BTW.

jpo38 Over a year ago

Sorry about that. access_key operator returns v.first, should it be v.second? Could you provide an example about how to use this access_key struct?

jpo38 Over a year ago

This gives access to the key, not the value, right? I adapted to value (changed v.first to v.second and return value from Key to Value but it does not work)

Ami Tavory Over a year ago

@jpos38 Just change first to second. The rest stays the same.

jpo38 Over a year ago

Let us continue this discussion in chat.

Yasin Yildirim · Accepted Answer · 2015-06-16 09:17:10Z

-2

One practical solution can be using extra boolean parameter to figure out whether container is map or not:

template <class Container, class Object> bool MyFindFunction( const Container& container, Object& found, bool isMap ){
   ..
   if(!isMap){
      found = *iter; // this works for set, but not for map
   }
   else{
      found = iter->second; // this works for map, but not for set
   }
..
}

edited Jun 16, 2015 at 9:17

answered Jun 16, 2015 at 9:10

Yasin Yildirim

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1 Comment

jpo38 Over a year ago

This does not work. Even if only used with isMap set to true, compiler complains error C2440: '=' : cannot convert from 'const std::pair<_Ty1,_Ty2>' to 'double'

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