[BUG] No implicit move for move this, explicit move doesn't parse

[JohelEGP]

Title: No implicit move for move this, explicit move doesn't parse.

Minimal reproducer (https://cpp2.godbolt.org/z/WhxKGfn83):

t: @basic_value type = {
  s: std::string = ();
  private get: (forward self) -> forward _ = self.s;
  operator*: (this) -> forward std::string = get(this);
  operator*: (move this) -> std::string = get(this); // `get(move this)` doesn't parse.
}
main: () = { }

Commands:

cppfront main.cpp2
clang++18 -std=c++23 -stdlib=libc++ -lc++abi -pedantic-errors -Wall -Wextra -Wconversion -Werror=unused-result -Werror=unused-value -Werror=unused-parameter -I . main.cpp

Expected result:

  [[nodiscard]] auto t::operator*() && -> std::string { return get(std::move(*this));  }

Actual result and error:

A copy.

  [[nodiscard]] auto t::operator*() && -> std::string { return get((*this));  }

Cpp2 lowered to Cpp1:

//=== Cpp2 type declarations ====================================================


#include "cpp2util.h"

#line 1 "/app/example.cpp2"
class t;
#line 2 "/app/example.cpp2"
  

//=== Cpp2 type definitions and function declarations ===========================

#line 1 "/app/example.cpp2"
class t {
#line 2 "/app/example.cpp2"
  private: std::string s {}; 
  private: [[nodiscard]] static auto get(auto&& self) -> auto&&;
  public: [[nodiscard]] auto operator*() const& -> std::string const&;
  public: [[nodiscard]] auto operator*() && -> std::string;
  public: t(t const& that);

public: auto operator=(t const& that) -> t& ;
public: t(t&& that) noexcept;
public: auto operator=(t&& that) noexcept -> t& ;
public: explicit t();

#line 6 "/app/example.cpp2"
};
auto main() -> int;

//=== Cpp2 function definitions =================================================

#line 1 "/app/example.cpp2"

#line 3 "/app/example.cpp2"
  [[nodiscard]] auto t::get(auto&& self) -> auto&& { return CPP2_FORWARD(self).s;  }
  [[nodiscard]] auto t::operator*() const& -> std::string const& { return get((*this));  }
  [[nodiscard]] auto t::operator*() && -> std::string { return get((*this));  }

  t::t(t const& that)
                                : s{ that.s }{}

auto t::operator=(t const& that) -> t& {
                                s = that.s;
                                return *this;}
t::t(t&& that) noexcept
                                : s{ std::move(that).s }{}
auto t::operator=(t&& that) noexcept -> t& {
                                s = std::move(that).s;
                                return *this;}
t::t(){}
#line 5 "/app/example.cpp2"
                                                     // `get(move this)` doesn't parse.

#line 7 "/app/example.cpp2"
auto main() -> int{}

Output:

Program returned: 0

See also:

• https://github.com/hsutter/cppfront/issues?q=is%3Aissue+is%3Aopen+move+in%3Atitle

[JohelEGP]

This does parse (https://cpp2.godbolt.org/z/3zdhbPnMo):

operator*: (move this) -> std::string = { return get(move this); }

I have other report(s) where f: () x; doesn't quite behave like f: () -> _ { return x; }:

• [BUG] Capture considered local variable #850.
• [BUG] Function initialized with == can't have _requires-clause_ #709.

[JohelEGP]

I think it's reasonable to also expect std::move on the implicit this. when accessing a member (https://cpp2.godbolt.org/z/3EnTfjhb6):

  operator*: (move this) -> std::string = s;

Expected:

  [[nodiscard]] auto t::operator*() && -> std::string { return std::move(s);  }

Actual:

  [[nodiscard]] auto t::operator*() && -> std::string { return s;  }

[JohelEGP]

Of course, the same should also apply to destructors (https://cpp2.godbolt.org/z/hvxxdaozq):

consume: (copy _: std::unique_ptr<int>) = { }
some_int: type = {
  value: std::unique_ptr<int>;
  operator=: (move this) = {
    consume(value); // Error: Can't copy. Move it!
  }
}
main: () = { }

[JohelEGP]

What to do for a data member of reference type?
x, where x is a data member, should lower to std::move(*this).x.
That "moves" x if it doesn't have a reference type.
If x has a reference type, the user has to explicitly move it.
If we moved a data member of reference type on last use,
the opt-out would look something like as-lvalue(x).

[JohelEGP]

_ = x; is not a good alternative to prevent the implicit move on last use of a reference member.
Generic code that handles both reference and non-reference data members benefit the most from moving *this and not x.
Otherwise, generic code can't be shared in a move this function that handles both kinds of members.

f: (forward _) = { }
t: @struct <T> type = {
  v: T;
  g: (move this) = { f(v); }      // If `v` lowers to `std::move(*this).x`, this can be correct in both cases.
  h: (move this) = { f(move v); } // If `v` lowers to `std::move(*this).x`, this can be correct in both cases.

  i: (move this) = { f(v); } // If `v` lowers to `std::move(x)`, this can be correct for `T = int`, but not for `T = int&`.

  j: (move this) = { f(v); } // If `v` lowers to `std::move(x)`, this can be correct for `T = int`,
  j: (move this)             // but if it isn't for `T = int&`, this overload is necessary.
    requires std::is_reference_v<T> = {
    f(v);
    _ = v;
  }
}

[JohelEGP]

that parameters already work as expected (https://cpp2.godbolt.org/z/4PvTahqb6):

f_copy: (copy _...) = { }
issue_857: @struct type = {
  i: std::unique_ptr<int>;
  f: (move this, move that) = f_copy(/*this,*/ that);
  g: (move this, move that) = f_copy(/*this.i,*/ that.i);
}