From 58eb921199d36c8fbe994a034f2cf6197120f90f Mon Sep 17 00:00:00 2001 From: Ralf Jung Date: Mon, 15 Jul 2019 13:32:33 +0200 Subject: [PATCH] mention valgrind connection --- personal/_posts/2019-07-14-uninit.md | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) diff --git a/personal/_posts/2019-07-14-uninit.md b/personal/_posts/2019-07-14-uninit.md index 72b26c7..314ca95 100644 --- a/personal/_posts/2019-07-14-uninit.md +++ b/personal/_posts/2019-07-14-uninit.md @@ -107,8 +107,11 @@ But *the Rust program you wrote does not run on your hardware*. It runs on the Rust abstract machine, and that machine (which only exists in our minds) *does* have a notion of "uninitialized memory". The real, physical hardware that we end up running the compiled program on is a very efficient *but imprecise* implementation of this abstract machine, and all the rules that Rust has for undefined behavior work together to make sure that this imprecision is not visible for *well-behaved* (UB-free) programs. But for programs that do have UB, this "illusion" breaks down, and [anything is possible](https://raphlinus.github.io/programming/rust/2018/08/17/undefined-behavior.html). + UB-free programs can be made sense of by looking at their assembly, but *whether* a program has UB is impossible to tell on that level. -For that, you need to think in terms of the abstract machine. +For that, you need to think in terms of the abstract machine.[^sanitizer] + +[^sanitizer]: This does imply that tools like valgrind, that work on the final assembly, can never reliably detect *all* UB. This does not just apply to uninitialized memory: for example, in x86 assembly, there is no difference between "relaxed" and "release"/"acquire"-style atomic memory accesses. But when writing Rust programs, even when writing Rust programs that you only intend to compile to x86, "what the hardware does" just does not matter. -- 2.30.2