Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

std::map::size_type for a std::map whose value_type is its own size_type

I have a std::map<std::pair<std::string, std::string>, float> that is taking up too much memory, and in order to use less memory, I've decided to map the unique strings to integers (e.g., std::map<std::string, int>, where each new unique string is mapped to the current size() of the map), and use those integer value as pairwise keys to the map, (e.g., std::map<std::pair<int, int>, float>).

Instead of int, I want to use std::map::size_type:

using map_index = std::map::size_type;
std::pair<map_index, map_index> key;

Of course, this doesn't compile because I need to supply the argument list for the map:

vector.cc:14:19: error: invalid use of template-name `std::map' without an argument list
 using map_index = std::map::size_type;

And this (in theory) is what I'm trying to achieve:

using map_index = std::map<std::string, map_index>::size_type;

which gives the following (expected) compiler error:

vector.cc:15:41: error: `map_index' was not declared in this scope
 using map_index = std::map<std::string, map_index>::size_type;

What is the proper way to get the compiler to infer the correct value_type for a std::map whose value_type is its own size_type?

like image 357
vallismortis Avatar asked Dec 02 '22 10:12

vallismortis


1 Answers

size_t should be good enough for such case.

But if you insist, you can do like this:

#include <type_traits>
#include <map>

template <class Key, class Value = size_t, size_t depth = 0, class = void>
struct GetSizeType {
    using type = typename GetSizeType<Key, typename std::map<Key, Value>::size_type, depth + 1>::type;
};

template <class Key, class Value, size_t depth>
struct GetSizeType<Key, Value, depth, std::enable_if_t<std::is_same_v<Value, typename std::map<Key, Value>::size_type>>> {
    using type = typename std::map<Key, Value>::size_type;
};

template <class Key, class Value>
struct GetSizeType<Key, Value, 100, void> {};

int main() {
    using X = GetSizeType<int>::type;

    return 0;
}

It will run recursively on GetSizeType, the recursive call will stop upon

  • reaching the recursive call depth limitation (there will be no member type in this case), or
  • finding a specialization of std::map of which the mapped_type and size_type is identical (the member type aliases size_type).
like image 154
felix Avatar answered Dec 11 '22 10:12

felix