Tukukkk

When restructuring some code I came across a 'problem' when returning a struct with 2 values. Now these really should be named for the documented effect. Later on I wanted to use tie so i changed the struct into inheriting from std::pair and just setting references. Now this actually works fine, but you will notice now my struct has the size of 24 as opposed to just 8 compared to the pair.

#include <tuple>





struct Transaction : public std::pair<int, int> {

    using pair::pair;



  int& deducted = first;

  int& transfered = second;

};

//static_assert(sizeof(Transaction) == sizeof(std::pair<int, int>));//commenting in this line will fail compilation



Transaction makeTheTransaction();



void test(int& deduct, int& transfer) {

    std::tie(deduct, transfer) = makeTheTransaction(); 

}

The maybe obvious method is to change into member functions, however that is also too much 'boilerplate' for this case (then it just becomes easier not use tie later on). A direct memcpy is to eg. a tuple is straigt forward UB. A direct structured binding is also not doable since the variables is already in use.

My question is what is a better or minimal code solution disregarding reusable parts (and given that the size shouldn't grow beyond the size of 2 ints) ?

UPDATE: For this case I ended up just doing a plain struct and hold return in a temporary. For other users coming here, there is a library proposal to boost that seems to be able to convert any struct to tuple: https://github.com/apolukhin/magic_get/

It seems like you're over-complicating the problem to me. If you need to use std::tie, you can just use it. There's no need to alter your structure:

struct Transaction

{

  int deducted;

  int transferred;

};



// later...



auto t = std::tie(transaction.deducted, transaction.transferred);

If this is a pattern you use frequently, then you can wrap it in a little helper method:

struct Transaction

{

  int deducted;

  int transferred;



  auto to_tuple() const

  {

    return std::tie(deducted, transferred);

  }

};

You can also use this to assign to multiple variables at once, although I strongly discourage that. It's error prone and leads to brittle code. (For example, if you reverse the order of deduct and transfer in the example below, you've got a bug, but the compiler will give no warning or error.)

void test(int& deduct, int& transfer)

{

  std::tie(deduct, transfer) = makeTheTransaction().to_tuple();

}

Edit: On second thought...

If the goal here is just easy decomposition of the struct into variables, you could do that directly and avoid using pairs or tuples:

struct Transaction

{

  int deducted;

  int transferred;



  void decompose(int* deducted_, int* transferred_)

  {

    *deducted_ = deducted;

    *transferred_ = transferred;

  }

};



void test(int& deduct, int& transfer)

{

  makeTheTransaction().decompose(&deduct, &transfer);

}

This is still brittle, but at least now you'll get intellisense when you write the call to the decompose method, which will make the pattern a little less error prone.

I would simply go with:

struct Transaction

{

    int deducted;

    int transfered;

};

With usage similar to:

Transaction makeTheTransaction() { return {4, 2}; }



int main()

{

    auto [deduct, transfer] = makeTheTransaction(); 

    std::cout << deduct << transfer << std::endl;

}

Demo

Adding a conversion function works:

#include <tuple>



struct Transaction {

    std::pair<int, int> data_;



    operator std::tuple<int &, int &> () {

        return std::tie(data_.first, data_.second);

    }

};

static_assert(sizeof(Transaction) == sizeof(std::pair<int, int>));



Transaction makeTheTransaction() {

    return Transaction();

}



void test(int& deduct, int& transfer) {

    std::tie(deduct, transfer) = makeTheTransaction();

}

I don't think that this will cause any lifetime issue when using with std::tie.

This Transaction doesn't work with structured binding with two identifiers. But you can make it supports "tuple-like" binding by specializing std::get and std::tuple_size for it.