CODEWITHSUNDEEP
syntaxrust
Nikolai Mavrenkov
Nikolai Mavrenkov

Reputation: 1923

Let &mut syntax

It is possible to make the following binding in Rust:

let &mut a = &mut 5;

But what does it mean exactly? For example, let a = &mut 5 creates an immutable binding of type &mut i32, let mut a = &mut 5 creates a mutable binding of type &mut i32. What about let &mut?

Upvotes: 13

Views: 2815

Answers (1)

Veedrac
Veedrac

Reputation: 60197

An easy way to test the type of something is to assign it to the wrong type:

let _: () = a;

In this case the value is an "integral variable", or a by-value integer. It is not mutable (as testing with a += 1 shows).

This is because you are using destructuring syntax. You are pattern matching your &mut 5 against an &mut _, much like if you wrote

match &mut 5 { &mut a => {
// rest of code
} };

Thus you are adding a mutable reference and immediately dereferencing it.

To bind a mutable reference to a value instead, you can do

let ref mut a = 5;

This is useful in destructuring to take references to multiple inner values.

Upvotes: 14

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