std::map default value

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;

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

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.

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.

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.

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;

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");

Pre-C++17, use std::map::insert(), for newer versions use try_emplace(). It may be counter-intuitive, but these functions effectively have the behaviour of operator[] with custom default values.

Realizing that I'm quite late to this party, but if you're interested in the behaviour of operator[] with custom defaults (that is: find the element with the given key, if it isn't present insert a chosen default value and return a reference to either the newly inserted value or the existing value), there is already a function available to you pre C++17: std::map::insert(). insert will not actually insert if the key already exists, but instead return an iterator to the existing value.

Say, you wanted a map of string-to-int and insert a default value of 42 if the key wasn't present yet:

std::map<std::string, int> answers;

int count_answers( const std::string &question)
{
    auto  &value = answers.insert( {question, 42}).first->second;
    return value++;
}

int main() {

    std::cout << count_answers( "Life, the universe and everything") << '\n';
    std::cout << count_answers( "Life, the universe and everything") << '\n';
    std::cout << count_answers( "Life, the universe and everything") << '\n';
    return 0;
}

which should output 42, 43 and 44.

If the cost of constructing the map value is high (if either copying/moving the key or the value type is expensive), this comes at a significant performance penalty, which would be circumvented with C++17's try_emplace().

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.

Expanding on the answer https://stackoverflow.com/a/2333816/272642, this template function uses std::map's key_type and mapped_type typedefs to deduce the type of key and def. This doesn't work with containers without these typedefs.

template <typename C>
typename C::mapped_type getWithDefault(const C& m, const typename C::key_type& key, const typename C::mapped_type& def) {
    typename C::const_iterator it = m.find(key);
    if (it == m.end())
        return def;
    return it->second;
}

This allows you to use

std::map<std::string, int*> m;
int* v = getWithDefault(m, "a", NULL);

without needing to cast the arguments like std::string("a"), (int*) NULL.

If you have access to C++17, my solution is as follows:

std::map<std::string, std::optional<int>> myNullables;
std::cout << myNullables["empty-key"].value_or(-1) << std::endl;

This allows you to specify a 'default value' at each use of the map. This may not necessarily be what you want or need, but I'll post it here for the sake of completeness. This solution lends itself well to a functional paradigm, as maps (and dictionaries) are often used with such a style anyway:

Map<String, int> myNullables;
print(myNullables["empty-key"] ?? -1);

With C++20 it is simple to write such getter:

constexpr auto &getOrDefault(const auto &map, const auto &key, const auto &defaultValue)
{
    const auto itr = map.find(key);
    return itr == map.cend() ? defaultValue : itr->second;
}

As @425nesp and КоеКто noticed, the compiler won't compile code with undefined behavior only if getOrDefault called as constant expression. If getOrDefault called as regular function, it possible to end up with a reference to the destroyed temporary.

Example

constexpr auto &v = getOrDefault(m, key, "foo"); // Compile error!
auto &v = getOrDefault(map, key, "foo"); // Undefined behavior!

const auto value = "foo";
auto &v = getOrDefault(map, key, value); // Ok

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;

Here is a correct approach that will conditionally return a reference if the caller passes in an lvalue reference to the mapped type.

template <typename Map, typename DefVal>
using get_default_return_t = std::conditional_t<std::is_same_v<std::decay_t<DefVal>,
    typename Map::mapped_type> && std::is_lvalue_reference_v<DefVal>,
    const typename Map::mapped_type&, typename Map::mapped_type>;

template <typename Map, typename Key, typename DefVal>
get_default_return_t<Map, DefVal> get_default(const Map& map, const Key& key, DefVal&& defval)
{
    auto i = map.find(key);
    return i != map.end() ? i->second : defval;
}

int main()
{
    std::map<std::string, std::string> map;
    const char cstr[] = "world";
    std::string str = "world";
    auto& ref = get_default(map, "hello", str);
    auto& ref2 = get_default(map, "hello", std::string{"world"}); // fails to compile
    auto& ref3 = get_default(map, "hello", cstr); // fails to compile
    return 0;
}

If you would like to keep using operator[] just like when you don't have to specify a default value other than what comes out from T() (where T is the value type), you can inherit T and specify a different default value in the constructor:

#include <iostream>
#include <map>
#include <string>

int main() {
  class string_with_my_default : public std::string {
  public:
    string_with_my_default() : std::string("my default") {}
  };

  std::map<std::string, string_with_my_default> m;

  std::cout << m["first-key"] << std::endl;
}

However, if T is a primitive type, try this:

#include <iostream>
#include <map>
#include <string>

template <int default_val>
class int_with_my_default {
private:
  int val = default_val;
public:
  operator int &() { return val; }
  int* operator &() { return &val; }
};

int main() {
  std::map<std::string, int_with_my_default<1> > m;

  std::cout << m["first-key"] << std::endl;
  ++ m["second-key"];
  std::cout << m["second-key"] << std::endl;
}

See also C++ Class wrapper around fundamental types