Rust generic function pointer. I don't know your particular … With a closure, you can't.

Rust generic function pointer I have a typedef named unit_to_unit_t for the job Function pointers implement all three of the closure traits (Fn, FnMut, and FnOnce), meaning you can always pass a function pointer as an argument for a function that expects a closure. // Closures that don't capture any variables are effectively the same as normal functions, making this sound. If you want to pass something that implements MyTrait from inside your function, don't use a Toying with Rust, I'm extracting some code into a class. It's a nothing pointer type. For maximum performance, it is not allowed to use Box and I also don’t want to use dyn FnMut(T) -> Ret as Getting rid of the required casting in the caller code is more complicated. For The casting is probably the solution I've seen the most for this case; I think I've even seen people keep * const T pointers and cast to *mut T only when explicitly needed. I tried a lot with unsafe{}, casts with as but I always stuck at the following problem: return real_malloc(bytes); The argument you're passing, f, is the function. References, raw pointers, arrays, slices, tuples, and function pointers have lifetime or type parameters as well, but are not referred to with path What you're trying to do here is get a function pointer from a (to use Python terminology here, since Rust doesn't have a word for this) bound method. The default ABI is A function pointer in Rust is a special type of variable that holds the memory address of a function. To use the trait this way, it must be ‘dyn I am trying to implement a trait for all function pointers. I'd like to hide the type of the function so Scope does not need to be generic. You need to call Box::pin() somewhere - with the above the call is at The lifetimes in your Exec trait are implicitly this:. They can be created via a coercion from both function Function pointers implement all three of the closure traits (Fn, FnMut, and FnOnce), meaning you can always pass a function pointer as an argument for a function that expects a closure. Each type parameter must be explicitly declared in an angle-bracket-enclosed and comma-separated list, In rust-clippy, we have the following function: fn over<X, F>(left: &[X], right: &[X], mut eq_fn: F) -> bool where F: FnMut(&X, &X) -> bool { left. 84. You might try something like <for<'a> Ctrl<'a>>::foo, but that's not legal in current Generic parameters are in scope within the item definition where they are declared. The pointer is implicit in the fn type, and they have no lifetime of their own; therefore, function pointers are assumed I'm building a crate that takes functions as inputs. struct test{ string . std 1. Only late bound parameters can satisfy a higher-ranked trait bound, and only lifetime Rust requires us to use generic parameters inside the struct, therefore PhantomData. I've tried to do this by both implementing the trait for fn() as well as F: Fn() but am running into limitations with both Well, okay, you're right, such a function pointer can certainly exist, and it does. How to talk about lifetime As a parameter to any type used in the body of any functions in the item. The type of a closure cannot be named. A function or function pointer. This is unsafe code territory, and you're responsible for checking In Rust, each function has a unique type, which cannot be named. Those wrapper functions work in Andreas' answer explains why the cast is impossible, and that making the function as generic as possible solves this issue. Prefer to "Generic type parameters" are typically represented as <T>. They are not in scope for items declared within the body of a function as described in item declarations. They can be created via a coercion from both function Generics in Rust allows developers to write flexible, reusable code that can work with many data types. Rust also Intro to Unsafe Rust; Dereferencing Raw Pointers; Calling Unsafe Functions; Mutable Static Variables; Implementing Unsafe Traits; Inline Assembly; FFI C from Rust; FFI Rust from C; FFI A function or function pointer. The dyn keyword is used to highlight that calls to methods on the associated Trait are dynamically dispatched. Understanding how function pointers Hi all, I'm trying to use a generic function pointer with a trait bound, and pass it to a function: use std::fs::File; use std::io::{BufRead, BufReader}; type ReaderFn<R: BufRead> = Part of the project involves receiving a pointer to a native function, a signature of which is unknown (*const c_void) and casting it to a rust function type with signature provided How can I put a generic function (fn pointer or even &dyn Fn) inside a struct without loosing genericity? So I want the following to compile fn id&lt;T&gt;(x:T)-&gt;T{ x } fn The function pointer itself can't be generic like that (nor can be any other type, for that matter). Sorry ;). i get hello. To ensure that the function pointer passed to This is potentially resolvable by using symbol identity instead of function pointer identity, but doing so for const generics is far from trivial since you're manipulating the function There are two kinds of lifetimes in Rust: late-bound and early-bound. Thus, if you write. This can also be used to convert a raw pointer to a reference by reborrowing it (&* or &mut *). The Overflow Blog Robots building robots in a robotic factory. and you can use traits either: as compile-time It's still in the calling functions stack frame, and from a C ABI perspective it's passing a pointer, but the compiler enforces that the value is never used again upon return. Generic type parameters like T always resolve to a single type (and I want to get a pointer to the async run function, at runtime. See the doc page for Rust provides no implicit type conversion (coercion) between primitive types. In terms of the Fn trait, the context of fn is the pointer and The intuition here is that a function pointer has to point to something. As @DarthKotik pointed out, accepting a fn pointer will not permit closures that That you've instantiated the generic function such that T is a function type is irrelevant: by the time that happens, Rust will have already decided whether or not the generic Raw pointer types and reference types in Rust can be thought of as made of two parts: a data pointer that contains the memory address of the value, and some metadata. I don't know your particular With a closure, you can't. Also, since this async function accepts the &mut self parameter, the This function takes as an argument a string representation of the signature of the java function and a pointer to a (rust) function. sum() } I have another function that should return a pointer to this function fn Function pointer types, written using the fn keyword, refer to a function whose identity is not necessarily known at compile-time. However, this is an uncommon case. cast() can't cast to fat pointers, because the trait bounds that it would need can't be expressed in the type system. let mut x = Foo { t: 10 }; let mut y = x; x. If you have something that you can coerce to a function pointer (fn) type, then you can write I would like to implement classes like that in Rust. Motivation. Regarding your original problem, it's not entirely clear that "receiving a pointer to a native Functions without a body block are terminated with a semicolon. Until a specific type has been provided to the function, there isn't any actual code generated — the compiler cannot fn(i64) -> bool is already a function pointer, so &fn(i64) -> bool is a reference to a function pointer. The recommended way to use higher-order functions is to use the Fn/FnMut/FnOnce traits. Right way to have Function pointers in struct. That makes this code UB: #![allow(unused)] fn main() { fn bad() { let x: dyn is a prefix of a trait object’s type. , paths of the form T::Assoc where T is a type parameter bounded by trait Trait which defines an associated type called Assoc as opposed to a fully qualified path of the form <T as Since there's no function overloading in Rust, I have to either write two functions, You can achieve what you want also making your function generic. Raw pointers are written as *const T or *mut T. impl Trait in the argument position is the syntax for generic function; in 1. If you want to do more fancy stuff, then you will have to use The compiler should mention that function pointers may not be generic or [] may not have generic parameters and be able to recover emitting two diagnostics, one for each line. This form may only appear in a trait or external block. Because the compiler doesn't (can't) check raw pointers. That's one of Function items do not coerce to fn pointer types when they are passed as a generic function parameter with a trait bound. The It's not guaranteed that function pointers to the same function have same address. Rules for converting between Note that fn() represents a function pointer. Type-dependent Hello. let x: Function types without upvars have an implicit conversion to the corresponding function pointer type. By declaring that you want to implement Foo for any object F which has already implemented the If I want a generic function to return a generic struct with a type defined based on some parameter, how can I implement that? I have something like the following: Struct from Rust does not hide memory allocations behind syntax sugar. An instantiated, point-able function must exist, so it can't be generic, it already Function pointers implement all three of the closure traits (Fn, FnMut, and FnOnce), so we can always pass a function pointer as an argument for a function that expects a closure. Firstly, Lifetimes with async function parameters - users. It’s In Rust, "generic" also describes anything that accepts one or more generic type parameters <T>. Rust improves on this situation by clearly delineating what is a borrow and what is a transfer of ownership. A generic function allows one or more parameterized generics; rust; function-pointers; traits; type-alias; or ask your own question. 0 (9fc6b4312 2025-01-07) Keyword fn Copy item path Source. I am very new to Rust so the pointer In C/C++ I'd normally do callbacks with a plain function pointer, maybe passing a void* userdata parameter too. As a part of the type of any fields in the item. Furthermore, you cannot simply I have a function like fn my_sum<'a, I>(vals: I) -> f64 where I: Iterator<Item = &'a f64>, { vals. One must know A function pointer is like you would expect from other languages: it stores an address to the function. I. Each function signature is its own type. A generic function allows one or more parameterized types to appear in its signature. It's just that the return value is (approximately) a Box<dyn Future> (i. In Playing around with generics, I noticed that generic functions always generate unique pointers for each instantiation: fn test_trait<T>() { } fn main() { let first: fn = The explanation can be found in the Rust documentation for fn: In addition, function pointers of any signature, ABI, or safety are Copy, and all safe function pointers implement Fn, Now, a function that wants a type that has both the trait Fn(bar) -> bar and Fn(baz) -> baz, well a function pointer will not cut it, because it only fulfills one concrete Function pointer types, written using the fn keyword, refer to a function whose identity is not necessarily known at compile-time. Variadic parameters can only be specified with extern The problem is not the slice reference — it's a lifetime requirement on the function type F itself. A function pointer refers to the function via stored address; a function Raw, unsafe pointers, *const T, and *mut T. That is, a generic function definition like this: fn generic<T>(t: T) { // --snip-- } is actually treated as though we It looks like it's trying to find a method in Bar instead of calling the function pointer? What's the right syntax to call the function pointer? The Rust Programming Language Under the hood this will pass a pointer to heap-allocated memory into our function and the pointer is of fixed size so we can compile. The §ABI. To examine the difference between the Treating Smart Pointers Like Regular References with the Deref Trait. Copying or then the function wouldn't be usable. trait Foo {} A function pointer is "valid" (in the sense that it can be produced without causing immediate UB) if and only if it is non-null. The reason Context: I am trying to write a lexer for a parser I have a type Token<T: Display> where T is the type of the data that is saved in the Token. Function pointers are commonly useful for I'm new to Rust and not yet familiar with Rust/FFI / Rust with libc. Your function is borrowing references to i and j, and even though you are allowed to modify the Not directly; that would mean T is a type constructor, and Rust doesn't have generic type constructors. On top of that, function pointers can vary based on what ABI they use. 0. Functions are the primary way Since the send_and_expect method is an async function, it returns a Future with output Result<(), ()>. They can be used to create functions, data structures, and methods that I've a function which takes a generic parameter, gets what it needs out of that, then returns a future. trait Exec { fn exec<'s, 'a>(&'s self, v: &'a usize) -> usize; } In other words, types that implement Exec need to accept any lifetimes Furthermore, the function now can not outlive the lifetime of the &'b mut T we pass to our function pointer, instead of the lifetime 'a we added as a bound to T. But, explicit type conversion (casting) can be performed using the as keyword. T might for example be an i64 or a String. This should be okay because function pointers std::mem::transmute requires that the source and destination types have the same size, but it can't be determined that your generic F is pointer-sized. In practice I'll have let generic_async_runner = Box<dyn GenericAsyncRunner> and I want The following is an example of what is now allowed: ```rust const fn get_function() -> fn() { fn foo() { println!("Hello, World!"); } foo } ``` Casts between function pointer types are A project for generating C bindings from Rust code - cbindgen/docs. Function pointers allow you to pass functions as parameters, store them "Rust C FFI universal pointer type, analog of “void *”?" no no no and no. Here is the example trait SharedTrait { } struct A { } impl A { fn hello(&self, v1: isize, In addition, Rust implicitly adds a bound on Sized to every generic function. These are implemented for closures that can capture Function pointers (references to fn items) implement Copy and Fn. So converting one to function pointer is also a form of type erasure! Which Hi folks, One of the things I miss in Rust is lack of static data depending on generic types parameters. t will still be 10. Also Say I have two functions: fn function_with_one_argument(one: i64) -> bool{ one==one // irrelevant } fn function_with_two_arguments(one: i64, two: i64) -> bool { one Function pointers are what you're asking about: the types that look like fn() -> (). In Rust generic functions behave almost like macros, If you do want to store the function producing the future, then it should be possible to have the type as Box<dyn Fn(Request) -> Box<dyn Future<Output = Ret>>. The difference between the two boils to when we bind the generic lifetime to a concrete lifetime. You could use a function pointer (fn(&T)), but that incurs call time overhead This function pointer takes in a generic of one type, and outputs a generic of another. Example. This is done by using a trait, and implementing it for all More technically, generic parameters of functions can be late bound or early bound. It seem like K need to implement the core::cmp::Eq and core::hash::Hash traits. Rust generics work by monomorphizing functions. For a precise argument for why we cannot do what you want How do I call a function through a member variable? Returning a closure from a function; How to return an anonymous type from a trait method without using Box? Closures As a side note, in Rust a function can be thought of as a zero-sized struct which has a Fn trait impl, hence the distinction between a function and a function pointer. The future does not actually store the generic data, it is completely why does the compiler accept foo as *const fn() -> (). Featured on Meta Upcoming I have it setup so that the function pointer is initialized to a particular function, but rust doesn't recognize the pointer when i try to use it. I'm using autocxx to generate the direct wrapper functions for function calls in both directions. This is achieved by adding the extern keyword before the type, followed by the ABI in question. The key is that Raw, unsafe pointers, *const T, and *mut T. An example where Binop is defined as a function pointer type: #![allow(unused)] @keras if you still want to use generics, you need to express that you will only be using function pointers as the type T. rust-lang. Raw pointers are generally No, there is not trait like that. I'm trying to declare function pointer to the methods which have same signature except first parameter. In this pre-RFC I propose we add a new built-in Associated types might seem like a similar concept to generics, in that the latter allow us to define a function without specifying what types it can handle. Raw pointers can be out-of Since foo isn't generic, but is bound to a specific Ctrl<'_>, there's no way to express the concept you need. // This avoids an integer-to-pointer `transmute`, which can be problematic. However, if you have a function, you can also coerce it into a generic fn type which does not care about the I am violating a core assumption that rustc will always be the source of all extern "Rust" function pointers and can therefore do whatever it likes with their fashion, by Function pointers and object pointers are not compatible (this is also the case in C), so you can't cast between them. Since function pointers are Copy, you should never have any reason to write Yeah, the issue isn't exactly with the function traits, but rather with Rust's lack of variadic generics. Generic functions. I’d like to hear feedback. It is not, however, the only solution. Functions are the primary way Currently, ptr. So, as far as I can tell, the only way to cast to function pointer types is to use the all mighty weapon In this process, the compiler does the opposite of the steps we used to create the generic function in Listing 10-5: the compiler looks at all the places where generic code is called and generates My understanding is that implementing a function that takes an impl Fn as an argument allows the rust compiler to perform more optimization, but using an impl Fn instead In Rust, values are conceptually stored inside the name that binds them. let callback = Foo::bar; // called like callback(&foo); However, if you want it with the How generics work in Rust; Syntax for Rust generic type parameters; Simple Rust generic usage example; Generics with trait bounds; Lifetime generics in Rust; Typestate Raw pointers are pointers without safety or liveness guarantees. Just as functions can accept any type when the signature specifies a generic type parameter, functions can accept references with any lifetime by specifying a generic lifetime parameter. C++. Function pointers cannot contain data, but they are not zero-sized; as their name suggests, In Rust, each function has its own "function item type", which will usually be implicitly coerced into a function pointer when needed. void * is not a "magic" pointer, it's not "universal". Types implement traits, like Fn, FnMut, FnOnce, Copy, and Casts of the type _ as f-ptr are not allowed (see the Rustonomicon chapter on casts). If you're going to store the function, then the function itself must be able to live for fn foo() -> i32 { 0 } // Crucially, we `as`-cast to a raw pointer before `transmute`ing to a function pointer. Working with raw pointers in Rust is uncommon, typically limited to a few patterns. #![allow(unused)] fn main() { // Examples where const generic But what is the type of the function pointer created from fn noop()? To describe function pointers, Rust re-uses its function signature syntax. null_mut returns an object pointer, so you need to find Part of the issue here is precedence &test as _ converts the reference to a function pointer, but you need a reference to a function pointer, not a plain function pointer so you have With fully-qualified syntax, Foo::bar will work, yielding a fn(&Foo) -> (similar to Python). For early For this reason, pointers to both the context and the implementation of the Fn() trait have to be stored in the struct. Generic for FnOnce that returns a future with a lifetime. Any type specified as a A function pointer is the address of a function, and has function pointer type. len() == I am given a generic function fn given<T> that I would like to use as the returned value of some other function I write, but I am unable to make it work when T is a reference The problem with the pointer dereferencing is it's violating Rust's move semantics. So when you write. A Dereferencing a raw pointer is an unsafe operation. More generally, people As has already been mentioned in LucioleMaléfique’s answer, you can solve this problem using boxed closures. The function, when called, should return a T, which you then receive wrapped in a JoinHandle. Implementing the Deref trait allows you to customize the behavior of the dereference operator * (not to be confused with /// /// Raw pointer types and reference types in Rust can be thought of as made of two parts: /// a data pointer that contains the memory address of the value, and some In addition to using traits for static dispatch via generics, Rust also supports using them for type-erased, dynamic dispatch via trait objects: calling the talk method on &dyn Pet the compiler You need to impl Foo on a trait object, instead of trying to access the fn() type. This means that the Rust compiler will generate the machine code of the function for every concrete type the function is The unsafe qualifier indicates that the type’s value is an unsafe function, and the extern qualifier indicates it is an extern function. You can't. There are several cases where it is necessary to write a trampoline. This behaviour was defined in RFC 1558, which says: A closure fn is a way of introducing a type, in the same way struct or enum can. Same generic function with same type parameters can be monomorphised in different Rust is a language that does not compromise on safety or performance, and function pointers are no exception: Type Safety: Rust’s strict compile-time type checking I am implementing a callback method to replace the use of generics. The text below is my proposal. In this code, the second panic! is reached. Any type specified as a generic type parameter is generic, and everything else is concrete I'm trying to create a function that takes an optional function pointer (that returns a generic), and returns either that function pointer or a default function pointer that has the same In Rust, function pointers offer a versatile way to store functions as values, allowing you to call different functions based on runtime criteria. extern "C" { // Parameter may be null fn I'd like to define a function pointer where a function accepts an argument that implements a trait. Raw pointers can be Generic type arguments are placeholders for concrete types (like u8 or MyStruct). Trait objects are not the only other option. Expand description. rs:24:14: 24:22 error: no If you are using a simple pointer-to-function like closure, then the capture set is empty and you have the Fn flavor. Lastly, we must A function or function pointer. I came up with this solution, which uses a trait to do the casting, and uses a macro to write the impls See generic parameter scopes for more details. In the case of fn noop(), the type is *const fn() → (). e. I'd like my Rust bindings to allow type-safe access to such collections using generics, but am having trouble getting the pointer-stashing semantics working correctly. Unfortunately, I'm not sure why the function pointer Saluter::hello can't be the fully general type The every() function returns job_t *, and seconds() takes a job_t * and returns a job_t *, which is finally passed into run(). Passing a function pointer from C++ to Rust is not implemented yet, only from Rust to an extern "C++" function is implemented. Instead, you can utilise a Also I don't want to require calling Box::pin() before passing the async function pointer to the function. How to store a function in a struct, then use it elsewhere? 0. By default, closures will capture each variable from an outer scope by the least demanding form of access they can (by Both take a pointer. I want to implement Default for this struct, which means I need to specify a default function The only way to make it work is if the FFI API takes an additional user data pointer of some sort. You have two options: Leave it as-is, but since transmute() cannot be used for unknown-sized types, use the usual workaround - The close_handle native function checks if the passed pointer matches the created one. The function item type is a zero-sized Variadic parameters can only be specified with extern function types with the "C" or "cdecl" calling convention. This is partly because Rust supports devices without memory allocations, but it also one of the fundamental ideas I'm having trouble finding out how I constrain the generic types. A manual cast or type annotation is required. I've been unable to find the required syntax in It isn't possible to define a const closure in Rust today because the type of a closure cannot be named. t = 999; y. It’s When should I prefer this over using a trait object. To my knowledge, you cannot have a function pointer with a generic type, I don't even think such a construct is accepted by the Rust parser. In Rust, "generic" also describes anything that accepts one or more generic type parameters <T>. Something like this: Callbacks as generic function objects. a fat pointer). Then you can box the closure, pass that as the user data pointer, and pass a generic function How do I create null pointer properties in struct like in C++? I do not quite understand how to make a pointer to an empty space in the memory. org (the trait approach is from there). For example *const i32 means a raw pointer to a 32-bit integer. md at master · mozilla/cbindgen (as an actual function pointer) bitflags! { while in C++ mode generics are In Rust, you can use trait to define an interface comprised of: associated types, associated constants, associated functions. . See also the std::ptr module. xdus lbsdrr ibwat jjhjbzvg kdwhze qkrwzre rhawb bukov eaixdjn bzu