std::map default value

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;
}

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;

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.