std::map default value - c++

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

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;

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;
}

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");

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.

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().

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);

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;

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;
}

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

Related

How can i get template's type if my instance is stl map?

I want a get_value template function.
please see the following code:
template<typename T>
(something i want) get_value(const T& m, int key) {
auto it = m.upper_bound(key);
return it == m.begin() ? (*it).second : (*--it).second; // please notice here,
// if my instance is map<int, map<string, int>> the return type should be m.second's type
// that's map<string, int>
}
int main() {
std::map<int, std::map<std::string, int>> m;
auto& it = get_value(m, 10);
}
as you can see, the template function should have a return type, which is depend on instance's second type, is there any method i can get this type to make the code runnable?
The "second type" in a std::map<K,V> is called std::map<K,V>::mapped_type. However, you can use auto to let the compiler deduce that for you:
template<typename T>
auto get_value(const T& m, int key) {
auto it = m.upper_bound(key);
return it == m.begin() ? (*it).second : (*--it).second; // please notice here,
}
or with explicit type:
template<typename T>
typename T::mapped_type get_value(const T& m, int key) {
auto it = m.upper_bound(key);
return it == m.begin() ? (*it).second : (*--it).second; // please notice here,
}
If you can use C++14 standard or above, the safest way to go would be to use decltype(auto) as the return type:
template<typename T>
decltype(auto) get_value(const T& m, int key);
The difference to a plain auto is that decltype(auto) preserves cv-qualifiers, and in your case you most likely want to forward exactly the same type that std::map gives you.
For example, since the actual code uses std::map<int, std::map<std::string, int>>, you might want to avoid copying the return value every time, and decltype(auto) will achieve that.

Function template and ambigous template parameter

I have several std::map<some_type, some_other_type> and I'm trying to write a function template Lookup as shown below.
The function template works fine when the key is a pointer or a scalar, but if the key is std::string there are problems.
#include <iostream>
#include <map>
// Usage :
// bool valueisinmap = Lookup(themap, thekey, thevalue);
template <typename TK, typename TV>
bool Lookup(std::map<TK, TV>& map, TK key, TV& value)
{
auto it = map.find(key);
if (it != map.end())
{
value = it->second;
return true;
}
else
{
return false;
}
}
int main()
{
std::map<std::string, std::string> m;
m.insert(std::make_pair("2", "two"));
std::string x;
std::string key = "2";
if (Lookup(m, key, x))
std::cout << "OK\n";
if (Lookup(m, "2", x)) // problem here
std::cout << "OK\n";
}
I understand why Lookup(m, "2", x) doesn't compile because the type of "2" is not std::string but is there a way to write the function template so I can use Lookup(m, "2", x) as well as Lookup(m, key, x), key being a std::string?
And if yes this raises a second question:
bool Lookup(std::map<TK, TV>& map, TK key, TV& value)
key is passed by value and if the type of key is std::string, a copy is made. Is there a way to pass key by reference (or some C++14 and plus magic) and still being able to use Lookup(m, "2", x)?
One way to solve this is to introduce a separate type parameter for the key type, as follows:
template <typename TKM, typename TK, typename TV>
bool Lookup(const std::map<TKM, TV>& map, const TK& key, TV& value)
{
auto it = map.find(key);
if (it != map.end())
{
value = it->second;
return true;
}
else
{
return false;
}
}
As long as TK is implicitly convertible to TKM, Lookup can be called with a key of type TK in conjunction with a map that has key type TKM.
You can introduce another template parameter for key as #Ton van den Heuvel answered, another way is to exclude it from template argument deduction:
template <typename TK, typename TV>
bool Lookup(std::map<TK, TV>& map, const std::type_identity_t<TK>& key, TV& value)
Then TK will be only deduced from the 1st parameter map; if you pass a const char[] to the function as key it'll be converted to std::string then passed as the argument. And you can make is pass-by-const-reference to avoid potential unnecessary copy.
LIVE
BTW: std::type_identity is supported from C++20; if your compiler doesn't support it, you can make your own easily.
template<typename T> struct type_identity { typedef T type; };
You need two things here. First, the key type can be deduced seaparately (typename K below). Second, you want to pass the key as const-qualified reference and setup the map with a C++14 transparent comparison function (typename Comp below) to avoid unnecessary copies (see this thread for details about transparent comparators).
template <typename TK, typename TV, typename Comp, typename K>
bool Lookup(const std::map<TK, TV, Comp>& map, const K& key, TV& value)
{
// Same as before...
}
std::map<std::string, std::string, std::less<>> m;
Specifying std::less<> as the std::map comparison type makes sure that overloads #3 and #4 of std::map::find are available.
Note that I have additionally const-qualified the map parameter itself, as the Lookup template doesn't modify it.

Const converting std containers

Consider that I have a std::vector.
std::vector<int> blah;
blah.push_back(1);
blah.push_back(2);
I now want to pass the vector somewhere and disallow modifying the contents of the objects its contains while still allowing to modify the container when able:
// Acceptable use:
void call_something() {
std::vector<int> blah;
blah.push_back(1);
blah.push_back(2);
// Currently, compiler error because of mismatching types
something(blah);
}
void something(std::vector<const int>& blah)
{
// Auto translates to 'const int'
for ( auto& i : blah ) {
// User cannot modify i.
std::cout << i << std::endl;
}
blah.push_back(blah.size()); // This should be acceptable
blah.emplace_back(); // This should be acceptable
return;
}
// Unacceptable use:
void something_else(const std::vector<int>& blah)
{
// Because of const vector, auto translates to 'const int'
for ( auto& i : blah ) {
std::cout << i std::endl;
}
blah.push_back(blah.size()); // This will present an unacceptable compiler error.
blah.emplace_back(); // This will present an unacceptable compiler error.
return;
}
Is there an easy way to do this?
To enable the operations you wish to allow while preventing the others, you need to take a fine-grained approach to your function's interface. For example, if your calling code were to pass const iterators (begin and end) as well as a back inserter (or custom back emplacer functor), then exactly the subset of operations you showed would be possible.
template <class Iter, class F>
void something(Iter begin, Iter end, F&& append)
{
using value_type = typename std::iterator_traits<Iter>::value_type;
std::copy(begin, end, std::ostream_iterator<value_type>(std::cout, "\n"));
append(std::distance(begin, end));
append();
return;
}
That said I don't find your examples particularly compelling. Do you have a real scenario in which you must maintain mutable elements, pass a mutable container to a function, yet treat the passed elements as immutable?
There is no easy way to do this. One way would be to wrap a vector in a type that exposes only the functionality that you want to allow. For instance
template<typename T, typename A = std::allocator<T>>
struct vector_wrap
{
using iterator = typename std::vector<T, A>::const_iterator;
using const_iterator = typename std::vector<T, A>::const_iterator;
using size_type = typename std::vector<T, A>::size_type;
vector_wrap(std::vector<T, A>& vec)
: vec_(&vec)
{}
void push_back(T const& value) { vec_->push_back(value); }
void push_back(T&& value) { vec_->push_back(std::move(value)); }
size_type size() { return vec_->size(); }
iterator begin() const { return vec_->cbegin(); }
iterator end() const { return vec_->cend(); }
private:
std::vector<T, A> *vec_;
};
Since the above implementation only stores a pointer to the vector it wraps, you'll have to ensure that the lifetime of the vector is longer than that of vector_wrap.
You'll have to modify something and something_else so that they take a vector_wrap<int> as argument. Since vector_wrap::begin and vector_wrap::end return const_iterators, you'll not be allowed to modify existing elements within the for statement.
Live demo

performance of emplace is worse than check followed by emplace

I have a std::unordered_map with a value_type that does not have a default constructor so I cannot do the following
auto k = get_key();
auto& v = my_map[k];
I ended up writing a helper function
value_type& get_value(key_type& key)
{
return std::get<0>(my_map.emplace(
std::piecewise_construct,
std::forward_as_tuple(key),
std::forward_as_tuple(args_to_construct_value)
))->second;
}
but the performance was markedly worse (i.e. the value_type's constructor showed up in perf) than the following version.
value_type& get_value(key_type& key)
{
auto it = my_map.find(key);
if (it == my_map.end())
return std::get<0>(my_map.emplace(
std::piecewise_construct,
std::forward_as_tuple(key),
std::forward_as_tuple(args_to_construct_value)
))->second;
else
return it->second;
}
I read from std::unordered_map::emplace object creation that emplace needs to construct the object in order to see if exists. But emplace is checking to see if this key value pair exists in the map before it returns.
Am I using emplace the wrong way? Is there a better pattern I should follow that:
Won't construct my value_type every lookup (as in my first method)
Won't do the check for to see if value_type exists in my map twice (as in my second method)
Thanks
Your code is unfortunately optimal for the standard library as it currently is.
The problem is that the emplace operation is designed to avoid copying, not to avoid unnecessary construction of the mapped type. In practical terms, what happens is that the implementation allocates and constructs a node, containing the map value_type i.e. pair<const Key, T>, and then hashes the key to determine whether the constructed node can be linked into the container; if this collides then the node is deleted.
As long as hash and equal_to are not too expensive, your code shouldn't do too much extra work.
An alternative is to use a custom allocator that intercepts 0-argument construction of your mapped type; the problem is that detecting such construction is pretty fiddly:
#include <unordered_map>
#include <iostream>
using Key = int;
struct D {
D() = delete;
D(D const&) = delete;
D(D&&) = delete;
D(std::string x, int y) { std::cout << "D(" << x << ", " << y << ")\n"; }
};
template<typename T>
struct A {
using value_type = T;
using pointer = T*;
using const_pointer = T const*;
using reference = T&;
using const_reference = T const&;
template<typename U> struct rebind { using other = A<U>; };
value_type* allocate(std::size_t n) { return std::allocator<T>().allocate(n); }
void deallocate(T* c, std::size_t n) { std::allocator<T>().deallocate(c, n); }
template<class C, class...Args> void construct(C* c, Args&&... args) { std::allocator<T>().construct(c, std::forward<Args>(args)...); }
template<class C> void destroy(C* c) { std::allocator<T>().destroy(c); }
std::string x; int y;
A(std::string x, int y): x(std::move(x)), y(y) {}
template<typename U> A(A<U> const& other): x(other.x), y(other.y) {}
template<class C, class...A> void construct(C* c, std::piecewise_construct_t pc, std::tuple<A...> a, std::tuple<>) {
::new((void*)c)C(pc, a, std::tie(x, y)); }
};
int main() {
using UM = std::unordered_map<Key, D, std::hash<Key>, std::equal_to<Key>, A<std::pair<const Key, D>>>;
UM um(0, UM::hasher(), UM::key_equal(), UM::allocator_type("hello", 42));
um[5];
}
You could use boost::optional<T> in order to be able to default construct the mapped type and then assign an initialized T to it later.
#include <cassert>
#include <unordered_map>
#include <boost/optional.hpp>
struct MappedType
{
explicit MappedType(int) {}
};
int main()
{
std::unordered_map<int, boost::optional<MappedType>> map;
boost::optional<MappedType>& opt = map[0];
assert(!opt.is_initialized());
opt = MappedType(2);
assert(opt.is_initialized());
MappedType& v = opt.get();
}
James, you've mostly answered your own question.
You're doing nothing wrong in either implementation. emplace simply does more work than find, especially when the key already exists in your unordered_map.
If your get_value helper function mostly receives duplicates, then calling emplace every time will cause a performance hot spot as you've observed.

Getting a list of values from a map

Is there an stl way to get a list of values from a map?
i.e, I have:
std::map<A,B> myMap;
and I would like a function that will return just the list of values, i.e, std::list<B> (or set for that matter.
Is there a built-in stl way to do this?
A map element is defined as a map::value_type, and the type of it is a pair<A,B>. first is the key and second is the value. You can write a functor to extract second from a value_type, and copy that in to a vector (or a list, or whatever you want.) The best way to do the copying is to use transform, which does just what its name implies: it takes a value of one type and transforms it to a different type of value.
Here's a complete working example:
#include <cstdlib>
#include <map>
#include <string>
#include <algorithm>
#include <iterator>
#include <vector>
#include <iostream>
using namespace std;
typedef map<unsigned, string> MyMap;
MyMap my_map;
struct get_second : public std::unary_function<MyMap::value_type, string>
{
string operator()(const MyMap::value_type& value) const
{
return value.second;
}
};
int main()
{
my_map[1] = "one";
my_map[2] = "two";
my_map[3] = "three";
my_map[4] = "four";
my_map[5] = "five";
// get a vector of values
vector<string> my_vals;
transform(my_map.begin(), my_map.end(), back_inserter(my_vals), get_second() );
// dump the list
copy( my_vals.begin(), my_vals.end(), ostream_iterator<string>(cout, "\n"));
}
EDIT:
If you have a compiler that supports C++0x lambdas, you can eliminate the functor entirely. This is very useful for making code more readable and, arguable, easier to maintain since you don't end up with dozens of little one-off functors floating around in your codebase. Here's how you would change the code above to use a lambda:
transform(my_map.begin(), my_map.end(), back_inserter(my_vals), [](const MyMap::value_type& val){return val.second;} );
There's nothing built in, no. It's simple enough to write your own function, though: Iterate over the map. The iterator will give you a pair<A, B>. Add each second value to the result list.
You can't just "get" such a list because there is no pre-existing list stored anywhere in the guts, but you can build one:
typedef std::map<A,B> myMapType;
myMapType myMap;
std::list<B> valueList;
for (myMapType::const_iterator it=myMap.begin(); it!=myMap.end(); ++it) {
valueList.push_back( it->second );
}
Or if you really like the more STL way:
class GetSecond {
template<typename T1, typename T2>
const T2& operator()( const std::pair<T1,T2>& key_val ) const
{ return key_val.second; }
};
typedef std::map<A,B> myMapType;
myMapType myMap;
std::list<B> valueList;
std::transform(myMap.begin(), myMap.end(), std::back_inserter(valueList),
GetSecond());
One of many "built-in" ways is of course the most obvious one. Just iterate over all pair elements, which are ordered by key (pair::first), and add the value (pair::second) to a new container, which you can construct with the correct capacity to get rid of excess allocations during the iteration and adding.
Just a note: std::list is seldom the container you actually want to be using. Unless, of course, you really, really do need its specific features.
Sure.
std::list<B> list;
std::for_each(myMap.begin(), myMap.end(), [&](const std::pair<const A, B>& ref) {
list.push_back(ref.second);
});
If you don't have a C++0x compiler, first you have my sympathies, and second, you will need to build a quick function object for this purpose.
You can use boost's transform_iterator: http://www.boost.org/doc/libs/1_64_0/libs/iterator/doc/transform_iterator.html
struct GetSecond {
template <typename K, typename T>
const T& operator()(const std::pair<K, T> & p) const { return p.second; }
template <typename K, typename T>
T& operator()(std::pair<K, T> & p) const { return p.second; }
};
template <typename MapType>
auto begin_values(MapType& m) -> decltype(boost::make_transform_iterator(m.begin(), GetSecond())) {
return boost::make_transform_iterator(m.begin(), GetSecond());
}
template <typename MapType>
auto end_values(MapType& m) -> decltype(boost::make_transform_iterator(m.end(), GetSecond())) {
return boost::make_transform_iterator(m.end(), GetSecond());
}
template <typename MapType>
struct MapValues {
MapType & m;
MapValues(MapType & m) : m(m) {}
typedef decltype(begin_values(m)) iterator;
iterator begin() { return begin_values(m); }
iterator end() { return end_values(m); }
};
template <typename MapType>
MapValues<MapType> get_values(MapType & m) {
return MapValues<MapType>(m);
}
int main() {
std::map<int, double> m;
m[0] = 1.0;
m[10] = 2.0;
for (auto& x : get_values(m)) {
std::cout << x << ',';
x += 1;
}
std::cout << std::endl;
const std::map<int, double> mm = m;
for (auto& x : get_values(mm)) {
std::cout << x << ',';
}
std::cout << std::endl;
}