std::map default value

Question

Is there a way to specify the default value std::map's operator[] returns when an key does not exist?

Answer 1

No, there isn't. The simplest solution is to write your own free template function to do this. Something like:

#include <string>
#include <map>
using namespace std;

template <typename K, typename V>
V GetWithDef(const  std::map <K,V> & m, const K & key, const V & defval ) {
   typename std::map<K,V>::const_iterator it = m.find( key );
   if ( it == m.end() ) {
      return defval;
   }
   else {
      return it->second;
   }
}

int main() {
   map <string,int> x;
   ...
   int i = GetWithDef( x, string("foo"), 42 );
}

---

C++11 Update

Purpose: Account for generic associative containers, as well as optional comparator and allocator parameters.

template <template<class,class,class...> class C, typename K, typename V, typename... Args>
V GetWithDef(const C<K,V,Args...>& m, K const& key, const V & defval)
{
    typename C<K,V,Args...>::const_iterator it = m.find( key );
    if (it == m.end())
        return defval;
    return it->second;
}

Answer 2

While this does not exactly answer the question, I have circumvented the problem with code like this:

struct IntDefaultedToMinusOne
{
    int i = -1;
};

std::map<std::string, IntDefaultedToMinusOne > mymap;

Answer 3

The C++ standard (23.3.1.2) specifies that the newly inserted value is default constructed, so map itself doesn't provide a way of doing it. Your choices are:

• Give the value type a default constructor that initialises it to the value you want, or
• Wrap the map in your own class that provides a default value and implements operator[] to insert that default.

Answer 4

template<typename T, T X>
struct Default {
    Default () : val(T(X)) {}
    Default (T const & val) : val(val) {}
    operator T & () { return val; }
    operator T const & () const { return val; }
    T val;
};

<...>

std::map<KeyType, Default<ValueType, DefaultValue> > mapping;

Answer 5

More General Version, Support C++98/03 and More Containers

Works with generic associative containers, the only template parameter is the container type itself.

Supported containers: std::map, std::multimap, std::unordered_map, std::unordered_multimap, wxHashMap, QMap, QMultiMap, QHash, QMultiHash, etc.

template<typename MAP>
const typename MAP::mapped_type& get_with_default(const MAP& m, 
                                             const typename MAP::key_type& key, 
                                             const typename MAP::mapped_type& defval)
{
    typename MAP::const_iterator it = m.find(key);
    if (it == m.end())
        return defval;

    return it->second;
}

Usage:

std::map<int, std::string> t;
t[1] = "one";
string s = get_with_default(t, 2, "unknown");

---

Here is a similar implementation by using a wrapper class, which is more similar to the method get() of dict type in Python: https://github.com/hltj/wxMEdit/blob/master/src/xm/xm_utils.hpp

template<typename MAP>
struct map_wrapper
{
    typedef typename MAP::key_type K;
    typedef typename MAP::mapped_type V;
    typedef typename MAP::const_iterator CIT;

    map_wrapper(const MAP& m) :m_map(m) {}

    const V& get(const K& key, const V& default_val) const
    {
        CIT it = m_map.find(key);
        if (it == m_map.end())
            return default_val;

        return it->second;
    }
private:
    const MAP& m_map;
};

template<typename MAP>
map_wrapper<MAP> wrap_map(const MAP& m)
{
    return map_wrapper<MAP>(m);
}

Usage:

std::map<int, std::string> t;
t[1] = "one";
string s = wrap_map(t).get(2, "unknown");

Answer 6

There is no way to specify the default value - it is always value constructed by the default (zero parameter constructor).

In fact operator[] probably does more than you expect as if a value does not exist for the given key in the map it will insert a new one with the value from the default constructor.

Answer 7

C++17 provides try_emplace which does exactly this. It takes a key and an argument list for the value constructor and returns a pair: an iterator and a bool.: http://en.cppreference.com/w/cpp/container/map/try_emplace

Answer 8

The value is initialized using the default constructor, as the other answers say. However, it is useful to add that in case of simple types (integral types such as int, float, pointer or POD (plan old data) types), the values are zero-initialized (or zeroed by value-initialization (which is effectively the same thing), depending on which version of C++ is used).

Anyway, the bottomline is, that maps with simple types will zero-initialize the new items automatically. So in some cases, there is no need to worry about explicitly specifying the default initial value.

std::map<int, char*> map;
typedef char *P;
char *p = map[123],
    *p1 = P(); // map uses the same construct inside, causes zero-initialization
assert(!p && !p1); // both will be 0

See Do the parentheses after the type name make a difference with new? for more details on the matter.

Answer 9

Maybe you can give a custom allocator who allocate with a default value you want.

template < class Key, class T, class Compare = less<Key>,
       class Allocator = allocator<pair<const Key,T> > > class map;

Answer 10

One workaround is to use map::at() instead of []. If a key does not exist, at throws an exception. Even nicer, this also works for vectors, and is thus suited for generic programming where you may swap the map with a vector.

Using a custom value for unregistered key may be dangerous since that custom value (like -1) may be processed further down in the code. With exceptions, it's easier to spot bugs.