Under the (false?) impression that boost::container::flat_set was a drop-in replacement of std::set , I

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Asked: June 14, 20262026-06-14T07:27:56+00:00 2026-06-14T07:27:56+00:00

Under the (false?) impression that boost::container::flat_set was a drop-in replacement of std::set , I

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Under the (false?) impression that boost::container::flat_set was a drop-in replacement of std::set, I replaced set with flat_set wherever I expected the number of element to be small and the search performance is more critical than inserts.

At a later stage, I was stumped by a confusing compilation error which I eventually traced to the use of flat_set as a class member.

For example:

class Room {
  private:
    boost::container::flat_set<int> v;
};

The following code will not compile, but works just fine if I replace flat_set with std::set.

Room a;
Room b = Room(); // Example 1. Compiles OK
a = b;           // Example 2. Compiles OK
a = Room();      // Example 3. Eeeek! Compile fails on this line

The compilation error I see is:

error: no match for ‘operator=’ in ‘a = Room()’
note: candidate is:
note: Room& Room::operator=(Room&)
note:   no known conversion for argument 1 from ‘Room’ to ‘Room&’

My questions are:

Is this error expected? If it is, then how do I work round it?
How are the three statements in the example code different, and why does only the last one fail?
Why does the compiler complain about the assignment operator of Room rather than flat_set? Has the use of flat_set influenced the default operators generated for the class?

Complete sample program:

#include <boost/container/flat_set.hpp>

class Room {
  private:
    boost::container::flat_set<int> v;
};

int main(int argc, char *argv[]) {
  Room a;

  Room b = Room();
  a = b;
  a = Room();  // compilation fails here

  return 0;
};