How do I make an unordered_map type have itself as a value? (C++)

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Apparently I'm getting rusty. I need to find a way to have an unordered_map reference a struct using just its forward declaration, but I want STL to handle allocation and cleanup like normal. See the code for a concrete example of what I want.

Basically I am trying to make an unordered_map derived type which has itself as a value member. I am trying to use unique_ptr to do so but not having access to make_ptr I'm at a loss. I'd like to use STL's automatic heap allocators and such for this. I just want the map to be able to have itself as a value, by hook or crook. Since I'm targeting the Arduino SDK (but for 32-bit cpus) I don't think I can use boost based solutions nor use C++14.

Again see the comments in the code I've provided for a clearer understanding of the question. I'll take any alternative that approximates what I want.

Thanks in advance

What I have tried:

C++

#include <memory>
#include <string>
#include <unordered_map>
struct map_value;
struct map_value {
  // what I *really* want - works with std::map on GCC i guess
  //std::unordered_map<std::string,map_value> value;
  
  // what I'll settle for
  std::unordered_map<std::string,std::unique_ptr<map_value>> value;
};
void setup() {
  
  map_value mv;
  map_value* pmv2 = new map_value();
  // what sorcery do i use below? i don't have std::make_unique nor can I get it 
  mv.value.insert(std::make_pair("test",pmv2));
}

void loop() {
}

Posted 10-Dec-20 5:58am

honey the codewitch

Updated 11-Dec-20 3:44am

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2 solutions

Solution 2

Forward declarations work only for pointers and references. Depending which compiler you using, the "map_value" is an incomplete type at the point where you declared it inside the struct map_value body. I beleive it probably should be:

C++

struct map_value;
struct map_value {
  // what I *really* want - works with std::map on GCC i guess
  //std::unordered_map<std::string,map_value> value;
  
  // what I'll settle for
  // std::unordered_map<std::string,std::unique_ptr<map_value>> value;
  std::unordered_map<std::string,std::unique_ptr<map_value*>> value;
};

Posted 11-Dec-20 3:44am

steveb

v2

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honey the codewitch 11-Dec-20 9:45am

unique_ptr takes the type and decorates it so you're creating a pointer to a pointer. Solution 1 is the correct answer.

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honey the codewitch · Accepted Answer · 2020-12-10T06:26:00

Try mv.value.insert(std::make_pair("test",std::unique_ptr<map_value>(pmv2))). std::make_unique is really simply... dynamic allocation (with new) following by constructing a std::unique_ptr to hold the pointer. The constructor is explicit so it won't convert from raw pointer to unique_ptr automatically, you have to ask it.

- Ben Voigt @ stackoverflow.com

I thought I had already tried this but apparently not.

How do I make an unordered_map type have itself as a value? (C++)

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Solution 2

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How do I make an unordered_map type have itself as a value? (C++)

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