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Understanding Rust Error E0614: Dereferencing a Non-dereferenceable Type

Introduction to Rust Error E0614

Rust Error E0614 occurs when attempting to dereference a variable of a type that does not implement the std::ops::Deref trait. This error is a compile-time error and can be fixed by ensuring the variable is of a dereferenceable type or by using a reference instead.

Erroneous example

Consider the following code snippet that leads to Rust Error E0614:

#![allow(unused)]
fn main() {
  let y = 0u32;
  *y; // error: type `u32` cannot be dereferenced
}

In this example, the variable y is of type u32 and does not implement std::ops::Deref, causing an error when attempting to dereference it using *y.

Fixing the error

To fix Rust Error E0614, ensure the variable is of a dereferenceable type or use a reference. Below is an example of how to fix this error:

#![allow(unused)]
fn main() {
  let y = 0u32;
  let x = &y;
  // `x` is a `&u32`, so we can dereference it:
  *x; // ok!
}

In this corrected example, we create a reference x to the variable y, allowing us to use dereferencing with *x.

Additional considerations

Implementing the std::ops::Deref trait on custom types allows for the custom type to be dereferenceable. Here's an example:

use std::ops::Deref;

struct MyType(u32);

impl Deref for MyType {
  type Target = u32;

  fn deref(&self) -> &u32 {
    &self.0
  }
}

fn main() {
  let my_value = MyType(42);
  *my_value; // ok, since MyType implements Deref
}

In this example, we implement the Deref trait for MyType, allowing instances of MyType to be dereferenced.

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