How to best access elements of std::pair? - c++

While using std::pair I came across two different approaches to access its elements. As they both seem to be valid and working I was wondering what is the difference between them and which approach is the preferred one?
std::pair<int, int> p(1,1); // can be of any type.
int i1 = p.first; // first approach
int i2 = std::get<0>(p); // second approach

If, in a given application, either of pair or 0 is not a literal but a parameter, use get:
template<class... T> auto sum0(const T&... t) {
return (std::get<0>(t)+...);
}
template<int i> auto sqrAt(const std::pair<int,double> &p) {
const auto v=std::get<i>(p);
return v*v;
}
If both pair and 0 are present literally, using .first is plainly preferable for readability reasons (including that it indicates the conscious use of std::pair):
template<class M>
void addKeys(M &m) {
for(auto &kv : m) kv.second+=kv.first;
}
Everything about this function indicates intended use with std::map or std::unordered_map, making it very readable despite the only type named being void.

Related

Comparer that takes the wanted attribute

In order to use a standard function like std::sort on some standard container Container<T>
struct T{
int x,y;
};
based on the y value, you need to write something like (for example):
std::vector<T> v;
//fill v
std::sort(v.begin(),v.end(),[](const auto& l,const auto& r){
return l.y<r.y;
});
The comparer that was written as lambda function is used too much and re-written again and again during the code for various classes and attributes.
Considering the case where y's type is comparable (either like int or there is an overload for the < operator), is there any way to achieve something like:
std::sort(v.begin(),v.end(),imaginary::less(T::y)); // Imaginary code
Is it possible in C++ to write such a function like less? or anything similar?
I am asking because I remember something like that in some managed language (I am not sure maybe C# or Java). However, I am not sure even about this information if it is true or not.
template<typename T, typename MT>
struct memberwise_less
{
MT T::* const mptr;
auto operator()(const T& left, const T& right) const
{ return (left.*mptr) < (right.*mptr); }
};
template<typename T, typename MT>
memberwise_less<T, MT> member_less(MT T::*mptr)
{
return { mptr };
}
and then you can do
std::sort(v.begin(), v.end(), member_less(&T::y));

Generic for loop for elementary and complex type

Suppose I have those two std::vector:
std::vector<int> v_int(1000);
std::vector<T> v_T(1000); // Where T is copy-costy type
if I need to loop through them (sepereatly) without the need for editing the items I may use:
for(const auto item:v_int){
//...
}
for(const auto& item:v_T){ //Note &
//...
}
Iterating using const auto item:v_T is too bad since a copy will be performed in each iteration. However, using const auto& item:v_int is not optimal but not that bad. So if I need a code that deal with both them I used to use const auto& item:v.
Question: Is there a generic way to write the for loop that will use the best declaration for both of them? Something like:
template <typename T>
void iterate(const std::vector<T> v){
for(const auto/*&*/ item:v){ // activate & if T is not elementary type
//...
}
}
You can do this using the standard type traits:
template <typename T>
using select_type = std::conditional_t<std::is_fundamental<T>::value,
const T, const T&>;
template <typename T>
void iterate(const std::vector<T> v){
for(select_type<T> item:v){ // activate & if T is not elementary type
//...
}
}
However, I question the wisdom of doing this. It just clutters the code for something which is unlikely to make any difference.
Use auto&& when you just don't care. Use auto const& when you must guarantee it is not edited.
References are aliases, and in an immediate context like a loop are trivial to optimize out of existence, especially if the value is never modified.
There are some difficulties when a reference is passed over a compilation unit boundary to a function, which is why passing int by value instead of const& is advised sometimes, but that does not apply here.
I am not a specialist in template area but I think what bothers you is that in case of char the copy will copy only one byte while using & will copy sizeof(char*) bytes.
What comes to my mind is to use the sizeof(T).
In this case you can use
if (sizeof(T) >= sizeof(void*))
// use '&'
else
// don't use '&'
since sizeof is evaluated in the compilation time, this if branch will be optimized out by the compiler.
Personally I would just use const auto& for every case because the overhead is negligible unless you're needing to crunch some very large data sets in tight loops.
An approach to obtain what you want however might be to implement template specializations for your iterate method:
#include <iostream>
#include <string>
#include <vector>
template <typename T>
void iterate(const std::vector<T>& v){
for(const auto& item:v){
std::cout << item << std::endl;
}
}
template <>
void iterate(const std::vector<int>& v){
for(const auto item:v){
std::cout << "[" << item << "]" << std::endl;
}
}
int main()
{
std::vector<std::string> strings = { "foo", "bar" };
std::vector<int> numbers = { 1, 2 };
iterate(strings);
iterate(numbers);
}
Note: you're more likely to see a performance hit from copy constructing your vectors into the iterate function :)

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

Large POD as tuple for sorting

I have a POD with about 30 members of various types and I will be wanting to store thousands of the PODs in a container, and then sort that container by one of those members.
For example:
struct Person{
int idNumber;
....many other members
}
Thousands of Person objects which I want to sort by idNumber or by any other member I choose to sort by.
I've been researching this for a while today and it seems the most efficient, or at least, simplest, solution to this is not use struct at all, and rather use tuple for which I can pass an index number to a custom comparison functor for use in std::sort. (An example on this page shows one way to implement this type of sort easily, but does so on a single member of a struct which would make templating this not so easy since you must refer to the member by name, rather than by index which the tuple provides.)
My two-part question on this approach is 1) Is it acceptable for a tuple to be fairly large, with dozens of members? and 2) Is there an equally elegant solution for continuing to use struct instead of tuple for this?
You can make a comparator that stores a pointer to member internaly so it knows which member to take for comparison:
struct POD {
int i;
char c;
float f;
long l;
double d;
short s;
};
template<typename C, typename T>
struct Comp {
explicit Comp(T C::* p) : ptr(p) {}
bool operator()(const POD& p1, const POD& p2) const
{
return p1.*ptr < p2.*ptr;
}
private:
T C::* ptr;
};
// helper function to make a comparator easily
template<typename C, typename T>
Comp<C,T> make_comp( T C::* p)
{
return Comp<C,T>(p);
}
int main()
{
std::vector<POD> v;
std::sort(v.begin(), v.end(), make_comp(&POD::i));
std::sort(v.begin(), v.end(), make_comp(&POD::d));
// etc...
}
To further generalize this, make make_comp take a custom comparator, so you can have greater-than and other comparisons.
1) Is it acceptable for a tuple to be fairly large, with dozens of members?
Yes it is acceptable. However it won't be easy to maintain since all you'll have to work with is an index within the tuple, which is very akin to a magic number. The best you could get is reintroduce a name-to-index mapping using an enum which is hardly maintainable either.
2) Is there an equally elegant solution for continuing to use struct instead of tuple for this?
You can easily write a template function to access a specific struct member (to be fair, I didn't put much effort into it, it's more a proof of concept than anything else so that you get an idea how it can be done):
template<typename T, typename R, R T::* M>
R get_member(T& o) {
return o.*M;
}
struct Foo {
int i;
bool j;
float k;
};
int main() {
Foo f = { 3, true, 3.14 };
std::cout << get_member<Foo, float, &Foo::k>(f) << std::endl;
return 0;
}
From there, it's just as easy to write a generic comparator which you can use at your leisure (I'll leave it to you as an exercise). This way you can still refer to your members by name, yet you don't need to write a separate comparator for each member.
You could use a template to extract the sort key:
struct A
{
std::string name;
int a, b;
};
template<class Struct, typename T, T Struct::*Member>
struct compare_member
{
bool operator()(const Struct& lh, const Struct& rh)
{
return lh.*Member < rh.*Member;
}
};
int main()
{
std::vector<A> values;
std::sort(begin(values), end(values), compare_member<A, int, &A::a>());
}
Maybe you want to have a look at boost::multi_index_container which is a very powerful container if you want to index (sort) object by different keys.
Create a class which can use a pointer to a Person member data to use for comparison:
std::sort(container.begin(), container.end(), Compare(&Person::idNumber));
Where Compare is:
template<typename PointerToMemberData>
struct Compare {
Compare(PointerToMemberData pointerToMemberData) :
pointerToMemberData(pointerToMemberData) {
}
template<typename Type
bool operator()(Type lhs, Type rhs) {
return lhs.*pointerToMemberData < rhs.*pointerToMemberData
}
PointerToMemberData pointerToMemberData;
};

std::map default value

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