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I don't think it's the case that if M is a free monad, then applying the free monad construction gets you back something isomorphic to M. So your question of "what is the free monad on X" is actually not related to "what functor is X the free monad of?". To show that monad M is not a free monad, the only thing we need to do is exhibit some equality that's true for M but not implied by the monad laws -- since the meaning of the free monad construction is that it guarantees the monad laws and nothing else.

Here's one way to do that for lists:

N3337 (~ C++11) and N4140 (~ C++14) say:

A trivially copyable class is a class that:

• has no non-trivial copy constructors,
• has no non-trivial move constructors,
• has no non-trivial copy assignment operators,
• has no non-trivial move assignment operators, and
• has a trivial destructor.

By this definition, S is trivially copyable.

N4659 (~ C++17) says:

A trivially copyable class is a class:

• where each copy constructor, move constructor, copy assignment operator, and move assignment operator is either deleted or trivial,
• that has at least one non-deleted copy constructor, move constructor, copy assignment operator, or move assignment operator, and
• that has a trivial, non-deleted destructor

By this definition, S is not trivially copyable.

N4860 (~ C++20) says:

A trivially copyable class is a class:

• that has at least one eligible copy constructor, move constructor, copy assignment operator, or move assignment operator,
• where each eligible copy constructor, move constructor, copy assignment operator, and move assignment operator is trivial, and
• that has a trivial, non-deleted destructor.

By this definition, S is not trivially copyable.

Thus, as published, S was trivally copyable in C++11 and C++14, but not in C++17 and C++20.

The change was adopted from DR 1734 in February 2016. Implementors generally treat DRs as though they apply to all prior language standards by convention. Thus, by the published standard for C++11 and C++14, S was trivially copyable, and by convention, newer compiler versions might choose to treat S as not trivially copyable in C++11 and C++14 modes. Thus, all compilers could be said to be correct for C++11 and C++14.

For C++17 and beyond, S is unambiguously not trivially copyable so GCC and Clang are incorrect. This is GCC bug #96288 and LLVM bug #39050

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try:

Only the memory consumption directly attributed to the object is accounted for, not the memory consumption of objects it refers to.

Since Decimal is a Python class with references to several other objects (EDIT: see below), you just get the total size of the references, which is constant — not including the referred values, which are not.

isn't that argument fatally flawed?

To the extent that the proof quoted from the paper is intended to demonstrate that a conforming C implementation is not permitted to perform the source-to-source transformation described in the question, or an equivalent, yes, the proof is flawed. The refutation presented in the question is sound.

There was some discussion in comments about how the refutation could be viewed as boiling down to anything being permissible in the event of undefined behavior. That is a valid perspective, and in no way does it undercut the argument. However, I think it's unnecessarily minimalistic.

Again, the key problem with the paper's proof is here:

the load of a can only return 0 (the initial value of a) because the store a=1 does not happen before it (because it is in a different thread that has not been synchronised with) and non-atomic loads must return the latest write that happens before them.

The proof's error is that the language specification's requirement that a read of a must return the result of a write to a that "happened before" it is conditioned on the program being free of data races. This is an essential foundation for the whole model, not some kind of escape hatch. The program manifestly is not free of data races if in fact the read of a is performed, so the requirement is moot in that case. The read of a by thread 2 absolutely can observe the write by thread 1, and there is good reason to suppose that it might sometimes do so in practice.

To look at it another way, the proof chooses to focus on the write not happening before the read, but ignores the fact that the read also does not happen before the write.

Taking the relaxed atomic accesses into account does not change anything. It is plausible that in a real execution of the paper's three-threaded program, the implementation (for example) speculatively executes the relaxed load of x in thread 2 on the assumption that it will return 1, then reads from a the value written by thread 1, and as a result, executes the store to y. Because the atomic accesses are performed with relaxed semantics, the execution of thread 3 can read the value of y as 1 (or speculate that it will do so) and consequently perform the write to x. All speculations involved can then be confirmed correct, with the final result that a = x = y = 1. It is intentional that this seemingly paradoxical result is allowed by the "relaxed" memory order.

isn't it indeed valid for a C11 implementation to treat the original three-threaded program as if it were the two-threaded program consisting of threads 2' and 3?

At minimum, the paper's argument does not show otherwise, even if we -- with no basis in the specification -- construe the scope of the UB arising from the data race to be limited to whether the value read from a is its initial one or the one written by thread 1.

Implementations are given broad license to behave as they choose, so long as they produce observable behavior that is consistent with the behavior required of the abstract machine. The creation and execution of multiple threads of execution is not itself part of the observable behavior of a program, as that is defined by the specification. Therefore, yes, a program that performed the proposed transformation and then behaved accordingly, or one that otherwise behaved as if there were a happens before edge between the write to a and the read from a, would not be acting inconsistently with the specification.