Rust code can be debugged using LLDB. The representation of variables used by tools like CodeLLDB, though, is simply a breakdown of the in-memory contents of the variable and does not show any information from the standard Debug trait.
Is there any way using LLDB I can invoke the actual Debug representation of a variable at runtime? This often has a significant amount of useful information which is not clear from a simple memory snapshot.
The "Debug" trait in Rust looks like a pretty close equivalent to the ObjC and Swift object description methods: it provides a to-string method that prints a developer-friendly view of the object. From what I can tell this trait cooperates with the standard formatted printing machinery in Rust.
If that's right, the natural way to give access to the Rust Debug Trait in lldb would be to implement the "Object Description" part of lldb's "Rust LanguageRuntime" and call the print function under the covers. This would be accessed by the po or print-object command in lldb.
Unfortunately, Rust doesn't have a "Rust LanguageRuntime" in lldb or really much of any support currently. The only mentions of Rust in the lldb sources are a recognizer for the Rust mangling scheme and a define that says "other than mangling, pretend Rust is C++". So that isn't a viable option at present.
You could also try calling Rust's print directly in the expression evaluator, but YMMV as calling Rust code in lldb doesn't always work: as it turns out, Rust is not C++...
I want to pass some params to rust linux kernel module. In C language, I can use module_param(). But I can't find a functional equivalent method in rust kernel.
I found similar method called kernel::miscdev::Registration::register_with_options(). But it needs a Options struct and that can only be created in kernel codes.
So, is there any way to pass parameters from cmd line to rust kernel module?
I'm currently in the progress of learning Rust. I'm mainly using The Rust Programming Language book and this nice reference which relates Rust features/syntax to C++ equivalents.
I'm having a hard time understanding where the core language stops and the standard library starts. I've encountered a lot of operators and/or traits which seems to have a special relationship with the compiler. For example, Rust has a trait (which from what I understand is like an interface) called Deref which let's a type implementing it be de-referenced using the * operator:
fn main() {
let x = 5;
let y = Box::new(x);
assert_eq!(5, x);
assert_eq!(5, *y);
}
Another example is the ? operator, which seems to depend on the Result and Option types.
Can code that uses those operators can be compiled without the standard library? And if not, what parts of the Rust language are depending on the standard library? Is it even possible to compile any Rust code without it?
The Rust standard library is in fact separated into three distinct crates:
core, which is the glue between the language and the standard library. All types, traits and functions required by the language are found in this crate. This includes operator traits (found in core::ops), the Future trait (used by async fn), and compiler intrinsics. The core crate does not have any dependencies, so you can always use it.
alloc, which contains types and traits related to or requiring dynamic memory allocation. This includes dynamically allocated types such as Box<T>, Vec<T> and String.
std, which contains the whole standard library, including things from core and alloc but also things with further requirements, such as file system access, networking, etc.
If your environment does not provide the functionality required by the std crate, you can choose to compile without it. If your environment also does not provide dynamic memory allocation, you can choose to compile without the alloc crate as well. This option is useful for targets such as embedded systems or writing operating systems, where you usually won't have all of the things that the standard library usually requires.
You can use the #![no_std] attribute in the root of your crate to tell the compiler to compile without the standard library (only core). Many libraries also usually support "no-std" compilation (e.g. base64 and futures), where functionality may be restricted but it will work when compiling without the std crate.
DISCLAIMER: This is likely not the answer you're looking for. Consider reading the other answers about no_std, if you're trying to solve a problem. I suggest you only read on, if you're interested in trivia about the inner workings of Rust.
If you really want full control over the environment you use, it is possible to use Rust without the core library using the no_core attribute.
If you decide to do so, you will run into some problems, because the compiler is integrated with some items defined in core.
This integration works by applying the #[lang = "..."] attribute to those items, making them so called "lang items".
If you use no_core, you'll have to define your own lang items for the parts of the language you'll actually use.
For more information I suggest the following blog posts, which go into more detail on the topic of lang items and no_core:
Rust Tidbits: What Is a Lang Item?
Oxidizing the technical interview
So yes, in theory it is possible to run Rust code without any sort of standard library and supplied types, but only if you then supply the required types yourself.
Also this is not stable and will likely never be stabilized and it is generally not a recommended way of using Rust.
When you're not using std, you rely on core, which is a subset of the std library which is always (?) available. This is what's called a no_std environment, which is commonly used for some types of "embedded" programming. You can find more about no_std in the Rust Embedded book, including some guidance on how to get started with no_std programming.
I am trying to extract bits of code from an embedded rust example that does not compile. A lot of these old embedded examples don't compile because they use nightly and they quickly become broken and neglected.
let mut buffer : [u8; 2048] = [0;2048];
// some code to fill the buffer here
// say we want to split the buffer at position 300
let (request_buffer, response_buffer) = buffer.split_mut_at(300);
This example uses #![no_std] so there is no standard library to link to and must have compiled at some point so the function split_mut_at must have worked at some point. I am using IntelliJ rust AND Visual Studio Code as the IDE but neither IDE's can point me to the definition of the split_mut_at function. There is a minefield of crates and use statements in the example and there is no clear way to pin-point where some function comes without huge trial and error effort.
btw, split_at_mut can usually be found in std::string::String
Is there a rust command that tells you what crate a function belongs to in your project? It always takes so long to update rust-docs when doing a rust update. Surely that can help!
You're looking for slice::split_at_mut (note the mut at the end). It is listed in the nightly docs here and the stable docs here. It is also indeed available with #![no_std]. It is defined in libcore here.
As a general rule of thumb when a function x from core or std has a mutable and immutable variant, the function requiring a immutable reference is named x and the function requiring a mutable reference is named x_mut.
I'm essentially looking for malloc/free in Rust.
I found alloc::heap, but when I try to use it I get the following error:
main.rs:1:1: 1:19 error: unstable feature
main.rs:1 ![feature(alloc)]
note: this feature may not be used in the stable release channel
If you want to do it in high-level Rust (RAII, memory safety, bounds checking), you should allocate by creating a Vec. You can optionally use into_boxed_slice to disable resizing the container. See the book's FFI example (specifically the uncompress function) for an example of using Vec as an allocator.
Either use a nightly Rust or you will have to use libc::funcs::c95::stdlib::malloc(size: size_t) from the libc crate. You'll have to transmute the result of course.