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charset=utf-8 Content-Transfer-Encoding: quoted-printable Received-SPF: pass client-ip=2a00:1450:4864:20::32c; envelope-from=zimon.toutoune@gmail.com; helo=mail-wm1-x32c.google.com X-Spam_score_int: -20 X-Spam_score: -2.1 X-Spam_bar: -- X-Spam_report: (-2.1 / 5.0 requ) BAYES_00=-1.9, DKIM_SIGNED=0.1, DKIM_VALID=-0.1, DKIM_VALID_AU=-0.1, DKIM_VALID_EF=-0.1, FREEMAIL_FROM=0.001, RCVD_IN_DNSWL_NONE=-0.0001, SPF_HELO_NONE=0.001, SPF_PASS=-0.001 autolearn=ham autolearn_force=no X-Spam_action: no action X-BeenThere: guix-devel@gnu.org X-Mailman-Version: 2.1.29 Precedence: list List-Id: "Development of GNU Guix and the GNU System distribution." List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , Errors-To: guix-devel-bounces+larch=yhetil.org@gnu.org Sender: guix-devel-bounces+larch=yhetil.org@gnu.org X-Migadu-Country: US X-Migadu-Flow: FLOW_IN X-Migadu-Queue-Id: 8CFF926A0B X-Migadu-Scanner: mx12.migadu.com X-Migadu-Spam-Score: -9.70 X-Spam-Score: -9.70 X-TUID: 2d/wSc1bbm4C Hi, On lun., 13 mai 2024 at 17:11, Richard Sent w= rote: > Instead of A and B building C directly, A and B download the > substitutable Guix package D, then use D to build C. Because D is a > reproducible package, it should be substitutable from both A and B. > Then, because D->C is the same for everyone, we could substitute the > Guix derivation computation for C. Maybe I am missing some details. From my understanding, your D is the result of the =E2=80=9CComputing derivation=E2=80=9D dance. And it is a mi= nimalist build because it must work for both cases: with or without substitutes. Somehow, my understanding of the current process is: A -> D -> C B -> D* -> C And, D or D* can be the same script. Here, the property is a kind of idempotency. Hence, C is substitutable. IIUC, your proposal is to fix D (the intermediary step). You propose the package named =E2=80=99guix=E2=80=99 that changes barely, but it could = be something more minimalist. The requirements is: susbtitutable. The problem is transferred to the first arrow, no? How can we be sure that A and B points to the same D? Other said, A lives in the past. Later, we changed D becoming D* because some bugs. The other revision B lives after this change. Therefore, we have the same picture: A -> D B -> D* But then how do we go to C since D and D* does not have a kind of idempotent property. >From my understanding, the current situation/process is very similar with the one for compiling compiler: you have your new Source of compiler and an OldCompiler binary, so you run: OldCompiler(Source) -> Compiler and again Compiler(Source) -> NewCompiler Somehow, whatever the choice of some OldCompiler binary, you get the same NewCompiler binary (aside trusting trust attack and friends, obviously ;-)) The story of =E2=80=9Cguix pull=E2=80=9D is not so different; from my under= standing. Again, maybe I am missing something. Cheers, simon