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Understanding Rust Error E0561

Introduction to Rust Error E0561

Rust Error E0561 occurs when a non-ident or non-wildcard pattern is used as a parameter of a function pointer type. This error is raised because function pointer types, unlike closures, have a specific type signature and don't accept patterns as function parameters. This article will delve into the details of Rust Error E0561, looking at examples and ways to resolve the error.

Erroneous Code Example

#![allow(unused)]
fn main() {
    type A1 = fn(mut param: u8); // error!
    type A2 = fn(&param: u32); // error!
}

In this example, the compiler raises Rust Error E0561 due to the usage of a mutable parameter (mut param: u8) and a reference parameter (&param: u32) in the function pointer types 'A1' and 'A2' respectively.

Resolving Rust Error E0561

To fix Rust Error E0561, remove the patterns from the function pointer types, retaining only identifiers or the wildcard symbol _ as function parameters. Here's the corrected version of the erroneous code example:

#![allow(unused)]
fn main() {
    type A1 = fn(param: u8); // ok!
    type A2 = fn(_: u32); // ok!
}

As you can see, we have removed the 'mut' keyword in 'A1' and the reference symbol '&' in 'A2'. The compiler now accepts the updated code without any errors.

Omitting Parameter Names

In some cases, you can also omit the parameter names altogether. This makes the function pointer type less readable but can be useful when the parameter's role is evident from the context. The following example demonstrates this approach:

#![allow(unused)]
fn main() {
    type A3 = fn(i16); // ok!
}

In this example, we've defined a function pointer type 'A3' without specifying any parameter name. Since Rust Error E0561 concerns improper usage of patterns in function pointer types, this code works without any issues.

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