Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

The std::transform-like function that returns transformed container

I'm trying to implement a function similar to std::transform algorithm but instead of taking the output iterator by an argument I want to create and return a container with transformed input elements.

Let's say that it's named transform_container and takes two arguments: container and functor. It should return the same container type but possibly parametrized by a different element type (the Functor can return element of different type).

I'd like to use my function as in the example below:

std::vector<int> vi{ 1, 2, 3, 4, 5 }; auto vs = transform_container(vi, [] (int i) { return std::to_string(i); });  //vs will be std::vector<std::string> assert(vs == std::vector<std::string>({"1", "2", "3", "4", "5"}));  std::set<int> si{ 5, 10, 15 }; auto sd = transform_container(si, [] (int i) { return i / 2.; });  //sd will be of type std::set<double> assert(sd == std::set<double>({5/2., 10/2., 15/2.})); 

I was able two write two functions — one for std::set and one for std::vector — that seem to work properly. They are identical, except of the container typename. Their code is listed below.

template<typename T, typename Functor> auto transform_container(const std::vector<T> &v, Functor &&f) -> std::vector<decltype(f(*v.begin()))> {     std::vector<decltype(f(*v.begin()))> ret;     std::transform(std::begin(v), std::end(v), std::inserter(ret, ret.end()), f);     return ret; }  template<typename T, typename Functor> auto transform_container(const std::set<T> &v, Functor &&f) -> std::set<decltype(f(*v.begin()))> {     std::set<decltype(f(*v.begin()))> ret;     std::transform(std::begin(v), std::end(v), std::inserter(ret, ret.end()), f);     return ret; } 

However, when I attempted to merge them into a single general function that works with any container, I encountered numerous issues. The set and vector are class templates, so my function template must take a template template parameter. Moreover, set and vector templates have a different number of type parameters that needs to be properly adjusted.

What is the best way to generalize the two function templates above into a function that works with any compatible container type?

like image 611
Michał W. Urbańczyk Avatar asked May 26 '14 13:05

Michał W. Urbańczyk


People also ask

What does std transform return?

Return valueAn iterator pointing to the element that follows the last element written in the result sequence.

What does transform return c++?

The transform() function in C++ sequentially applies an operation to the elements of an array(s) and then stores the result in another output array. The transform function is used in two forms: Unary operation: The operation is applied to each element in the input range, and the result is stored in the output array.

How do you use std transform?

std::transform on a range For example, to obtain the keys that a map contains, you can use std::transform the following way: map<int, string> m = { {1,"foo"}, {42, "bar"}, {7, "baz"} }; vector<int> keys; std::transform(m. begin(), m. end(), std::back_inserter(keys), getFirst);

How do you use transform in CPP?

The transform function is present in the C++ STL. To use it, we have to include the algorithm header file. This is used to perform an operation on all elements. For an example if we want to perform square of each element of an array, and store it into other, then we can use the transform() function.


2 Answers

Simplest cases: matching container types

For the simple case where the input type matches the output type (which I've since realized is not what you're asking about) go one level higher. Instead of specifying the type T that your container uses, and trying to specialize on a vector<T>, etc., just specify the type of the container itself:

template <typename Container, typename Functor> Container transform_container(const Container& c, Functor &&f) {     Container ret;     std::transform(std::begin(c), std::end(c), std::inserter(ret, std::end(ret)), f);     return ret; } 

More complexity: compatible value types

Since you want to try to change the item type stored by the container, you'll need to use a template template parameter, and modify the T to that which the returned container uses.

template <     template <typename T, typename... Ts> class Container,     typename Functor,     typename T, // <-- This is the one we'll override in the return container     typename U = std::result_of<Functor(T)>::type,     typename... Ts > Container<U, Ts...> transform_container(const Container<T, Ts...>& c, Functor &&f) {     Container<U, Ts...> ret;     std::transform(std::begin(c), std::end(c), std::inserter(ret, std::end(ret)), f);     return ret; } 

What of incompatible value types?

This only gets us partway there. It works fine with a transform from signed to unsigned but, when resolving with T=int and U=std::string, and handling sets, it tries to instantiate std::set<std::string, std::less<int>, ...> and thus doesn't compile.

To fix this, we want to take an arbitrary set of parameters and replace instances of T with U, even if they are the parameters to other template parameters. Thus std::set<int, std::less<int>> should become std::set<std::string, std::less<std::string>>, and so forth. This involves some custom template meta programming, as suggested by other answers.

Template metaprogramming to the rescue

Let's create a template, name it replace_type, and have it convert T to U, and K<T> to K<U>. First let's handle the general case. If it's not a templated type, and it doesn't match T, its type shall remain K:

template <typename K, typename ...> struct replace_type { using type = K; }; 

Then a specialization. If it's not a templated type, and it does match T, its type shall become U:

template <typename T, typename U> struct replace_type<T, T, U> { using type = U; }; 

And finally a recursive step to handle parameters to templated types. For each type in a templated type's parameters, replace the types accordingly:

template <template <typename... Ks> class K, typename T, typename U, typename... Ks> struct replace_type<K<Ks...>, T, U>  {     using type = K<typename replace_type<Ks, T, U>::type ...>; }; 

And finally update transform_container to use replace_type:

template <     template <typename T, typename... Ts> class Container,     typename Functor,     typename T,     typename U = typename std::result_of<Functor(T)>::type,     typename... Ts,     typename Result = typename replace_type<Container<T, Ts...>, T, U>::type > Result transform_container(const Container<T, Ts...>& c, Functor &&f) {     Result ret;     std::transform(std::begin(c), std::end(c), std::inserter(ret, std::end(ret)), f);     return ret; } 

Is this complete?

The problem with this approach is it is not necessarily safe. If you're converting from Container<MyCustomType> to Container<SomethingElse>, it's likely fine. But when converting from Container<builtin_type> to Container<SomethingElse> it's plausible that another template parameter shouldn't be converted from builtin_type to SomethingElse. Furthermore, alternate containers like std::map or std::array bring more problems to the party.

Handling std::map and std::unordered_map isn't too bad. The primary problem is that replace_type needs to replace more types. Not only is there a T -> U replacement, but also a std::pair<T, T2> -> std::pair<U, U2> replacement. This increases the level of concern for unwanted type replacements as there's more than a single type in flight. That said, here's what I found to work; note that in testing I needed to specify the return type of the lambda function that transformed my map's pairs:

// map-like classes are harder. You have to replace both the key and the key-value pair types // Give a base case replacing a pair type to resolve ambiguities introduced below template <typename T1, typename T2, typename U1, typename U2> struct replace_type<std::pair<T1, T2>, std::pair<T1, T2>, std::pair<U1, U2>> {     using type = std::pair<U1, U2>; };  // Now the extended case that replaces T1->U1 and pair<T1,T2> -> pair<T2,U2> template <template <typename...> class K, typename T1, typename T2, typename U1, typename U2, typename... Ks> struct replace_type<K<T1, T2, Ks...>, std::pair<const T1, T2>, std::pair<const U1, U2>> {     using type = K<U1, U2,          typename replace_type<              typename replace_type<Ks, T1, U1>::type,             std::pair<const T1, T2>,             std::pair<const U1, U2>         >::type ...     >; }; 

What about std::array?

Handling std::array adds to the pain, as its template parameters cannot be deduced in the template above. As Jarod42 notes, this is due to its parameters including values instead of just types. I've gotten partway by adding specializations and introducing a helper contained_type that extracts T for me (side note, per Constructor this is better written as the much simpler typename Container::value_type and works for all types I've discussed here). Even without the std::array specializations this allows me to simplify my transform_container template to the following (this may be a win even without support for std::array):

template <typename T, size_t N, typename U> struct replace_type<std::array<T, N>, T, U> { using type = std::array<U, N>; };  // contained_type<C>::type is T when C is vector<T, ...>, set<T, ...>, or std::array<T, N>. // This is better written as typename C::value_type, but may be necessary for bad containers template <typename T, typename...> struct contained_type { };  template <template <typename ... Cs> class C, typename T, typename... Ts> struct contained_type<C<T, Ts...>> { using type = T; };  template <typename T, size_t N> struct contained_type<std::array<T, N>> { using type = T; };  template <     typename Container,     typename Functor,     typename T = typename contained_type<Container>::type,     typename U = typename std::result_of<Functor(T)>::type,     typename Result = typename replace_type<Container, T, U>::type > Result transform_container(const Container& c, Functor &&f) {     // as above } 

However the current implementation of transform_container uses std::inserter which does not work with std::array. While it's possible to make more specializations, I'm going to leave this as a template soup exercise for an interested reader. I would personally choose to live without support for std::array in most cases.

View the cumulative live example


Full disclosure: while this approach was influenced by Ali's quoting of Kerrek SB's answer, I didn't manage to get that to work in Visual Studio 2013, so I built the above alternative myself. Many thanks to parts of Kerrek SB's original answer are still necessary, as well as to prodding and encouragement from Constructor and Jarod42.

like image 169
Michael Urman Avatar answered Oct 14 '22 23:10

Michael Urman


Some remarks

The following method allows to transform containers of any type from the standard library (there is a problem with std::array, see below). The only requirement for the container is that it should use default std::allocator classes, std::less, std::equal_to and std::hash function objects. So we have 3 groups of containers from the standard library:

  1. Containers with one non-default template type parameter (type of value):

    • std::vector, std::deque, std::list, std::forward_list, [std::valarray]
    • std::queue, std::priority_queue, std::stack
    • std::set, std::unordered_set
  2. Containers with two non-default template type parameters (type of key and type of value):

    • std::map, std::multi_map, std::unordered_map, std::unordered_multimap
  3. Container with two non-default parameters: type parameter (type of value) and non-type parameter (size):

    • std::array

Implementation

convert_container helper class convert types of known input container type (InputContainer) and output value type (OutputType) to the type of the output container(typename convert_container<InputContainer, Output>::type):

template <class InputContainer, class OutputType> struct convert_container;  // conversion for the first group of standard containers template <template <class...> class C, class IT, class OT> struct convert_container<C<IT>, OT> {     using type = C<OT>; };  // conversion for the second group of standard containers template <template <class...> class C, class IK, class IT, class OK, class OT> struct convert_container<C<IK, IT>, std::pair<OK, OT>> {     using type = C<OK, OT>; };  // conversion for the third group of standard containers template     <         template <class, std::size_t> class C, std::size_t N, class IT, class OT     > struct convert_container<C<IT, N>, OT> {     using type = C<OT, N>; };  template <typename C, typename T> using convert_container_t = typename convert_container<C, T>::type; 

transform_container function implementation:

template     <         class InputContainer,         class Functor,         class InputType = typename InputContainer::value_type,         class OutputType = typename std::result_of<Functor(InputType)>::type,         class OutputContainer = convert_container_t<InputContainer, OutputType>     > OutputContainer transform_container(const InputContainer& ic, Functor f) {     OutputContainer oc;      std::transform(std::begin(ic), std::end(ic), std::inserter(oc, oc.end()), f);      return oc; } 

Example of use

See live example with the following conversions:

  • std::vector<int> -> std::vector<std::string>,
  • std::set<int> -> std::set<double>,
  • std::map<int, char> -> std::map<char, int>.

Problems

std::array<int, 3> -> std::array<double, 3> conversion doesn't compile because std::array haven't insert method which is needed due to std::inserter). transform_container function shouldn't also work for this reason with the following containers: std::forward_list, std::queue, std::priority_queue, std::stack, [std::valarray].

like image 30
Constructor Avatar answered Oct 14 '22 23:10

Constructor