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Bradlee Speice 2019-02-16 20:46:25 -05:00
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35
_posts/2019-02-14-insane-allocators.md
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@ -1,7 +1,7 @@
---
layout: post
title: "Insane Allocators: segfaults in safe Rust"
description: "\"Trusting trust\" with allocators."
description: "...and what it means to be \"safe.\""
category: rust, memory
tags: []
---
@ -63,16 +63,23 @@ pub extern "C" fn malloc(size: usize) -> *mut c_void {
// for all subsequent allocations, corrupting the location.
return ALLOC;
}
// Note that we don't ever handle `free`; if the first object
// we allocate gets freed, the memory address being given
// to everyone becomes a "use-after-free" bug.
}
```
Because this implementation of `malloc` is intentionally broken,
every program run using this library will crash. And I mean *every*
program; if you use dynamic memory, you're going down.
So how is it possible to run the Rust compiler in this environment?
`LD_PRELOAD` applies to all programs, so running `ls` will also
lead to memory corruption and crashing! The answer is that `sudo`
`LD_PRELOAD` applies to all programs, so the compiler should also
encounter memory corruption and crash, right? The answer is that `sudo`
deletes environment variables like `LD_PRELOAD` and
`LD_LIBRARY_PATH` when running commands; it's possible to
crash `sudo` in the same way by using our evil `malloc`
implementation.
`LD_LIBRARY_PATH` when running commands. While it is technically possible
to crash `sudo` in the same way using our evil `malloc` implementation,
the default policy is to delete these variables because of security concerns.
Finally, why does Rust 1.31 work, and 1.32 fail? The answer is in the
release notes:
@ -88,12 +95,12 @@ global allocator. Rust programs prior to 1.28 aren't subject to this
# So what?
It should be made very clear: the code demonstrated here isn't a
security issue. "Safe" Rust programs are only crashing in these
circumstances because the memory allocator is intentionally lying to it.
Even in mission critical systems, there are a lot of concerns beyond memory allocation; the
I do want to clarify: the code demonstrated here isn't a
security issue, and doesn't call into question Rust's definition of "safe."
The code demonstrated here crashes because the memory allocator is lying to it.
Even in mission critical systems, there are a lot of concerns beyond allocators; the
[F-35 Joint Strike Fighter coding standards](http://www.stroustrup.com/JSF-AV-rules.pdf)
don't even give it a full page.
give memory allocation about 10 sentences total.
But this example does highlight an assumption of Rust's memory model
that I haven't seen discussed much: **safe Rust is safe if, and only if,
@ -101,7 +108,9 @@ the allocator it relies on is "correct"**. And because writing a non-trivial all
[fundamentally unsafe](https://doc.rust-lang.org/std/alloc/trait.GlobalAlloc.html#unsafety),
safe Rust will always rely on unsafe Rust somewhere.
That all said, know that "safe" Rust can only claim to be safe because it stands
That all said, know that "safe" Rust can claim to be safe because it stands
on the shoulders of incredible developers working on jemalloc,
[kmalloc](https://linux-kernel-labs.github.io/master/labs/kernel_api.html#memory-allocation),
and others.
and others. Without being able to trust the allocators, we wouldn't
be able to trust the promise of safe Rust. So to all the people
who make the safety promises of Rust possible - thanks.