Begin a semi-formal argument for correctness

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Jannis Christopher Köhl 2022-08-31 17:43:53 +02:00
parent 292869493c
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@ -15,6 +15,34 @@
//! place. For example: Assume `let x = (0, 0)` and that we want to propagate values from `x.0` and
//! `x.1` also through the assignment `let y = &x`. In this case, we should register `x.0`, `x.1`,
//! `(*y).0` and `(*y).1`.
//!
//!
//! # Correctness
//!
//! Warning: This is a semi-formal attempt to argue for the correctness of this analysis. If you
//! find any weak spots, let me know! Recommended reading: Abstract Interpretation.
//!
//! In the following, we will assume a constant propagation analysis. Our analysis is correct if
//! every transfer function is correct. This is the case if for every pair (f, f#) and abstract
//! state s, we have f(y(s)) <= y(f#(s)), where s is a mapping from tracked place to top, bottom or
//! a constant. Since pointers (and mutable references) are not tracked, but can be used to change
//! values in the concrete domain, f# must assume that all places that can be affected in this way
//! for a given program point are marked with top (otherwise many assignments and function calls
//! would have no choice but to mark all tracked places with top). This leads us to an invariant:
//! For all possible program points where there could possibly exist a mutable reference or pointer
//! to a tracked place (in the concrete domain), this place must be assigned to top (in the
//! abstract domain). The concretization function y can be defined as expected for the constant
//! propagation analysis, although the concrete state of course contains all kinds of non-tracked
//! data. However, by the invariant above, no mutable references or pointers to tracked places that
//! are not marked with top may be introduced.
//!
//! Note that we (at least currently) do not differentiate between "this place may assume different
//! values" and "a pointer to this place escaped the analysis". However, we still want to handle
//! assignments to constants as usual for f#. This adds an assumption: Whenever we have an
//! assignment, all mutable access to the underlying place (which is not observed by the analysis)
//! must be invalidated. This is (hopefully) covered by Stacked Borrows.
//!
//! To be continued...
use std::fmt::{Debug, Formatter};