2020-08-27
1731
#rust
Thomas Heartman
23920
Aug 27, 2020 ⋅ 6 min read

Understanding lifetimes in Rust

Thomas Heartman Developer, speaker, musician, and fitness instructor.

Recent posts:

How Ralph makes Claude Code actually finish tasks

Claude Code is deceptively capable. Point it at a codebase, describe what you need, and it’ll autonomously navigate files, write […]

Ikeh Akinyemi
Feb 17, 2026 ⋅ 4 min read
ai dev tool power rankings

AI dev tool power rankings & comparison [Feb. 2026]

Compare the top AI development tools and models of February 2026. View updated rankings, feature breakdowns, and find the best fit for you.

Chizaram Ken
Feb 13, 2026 ⋅ 10 min read

How to solve package validation pain with Publint

Broken npm packages often fail due to small packaging mistakes. This guide shows how to use Publint to validate exports, entry points, and module formats before publishing.

Rahul Chhodde
Feb 12, 2026 ⋅ 5 min read
feb 11 the replay

The Replay (2/11/26): React performance wins, fine-grained frameworks, and more

Discover what’s new in The Replay, LogRocket’s newsletter for dev and engineering leaders, in the February 11th issue.

Matt MacCormack
Feb 11, 2026 ⋅ 34 sec read
View all posts

2 Replies to "Understanding lifetimes in Rust"

  1. You write: “Lifetimes are what the Rust compiler uses to keep track of how long references are valid for.” But what about keeping track of which objects are borrowed? If I have a function f with signature fn f(x: &’a i32) -> &’a i32; and I do let x = 0; let y = f(&x); then rust borrow checker will consider y to be borrowing x . I don’t get this.

  2. Hey! Thanks for the question. Let me try and answer it for you.

    > How does the compiler keep track of which objects are borrowed?

    Any reference is a borrow. Whenever you have a value that’s not the owned instance, you have a borrow. In other words, keeping track of borrows is the same as keeping track of references. Declaring references (and lifetimes) in function signatures helps the compiler get the information it needs to keep track of borrows.

    > Why is `y` borrowing `x`?

    In your example, the function `f` takes a reference and returns the same reference. You then assign `y` to that reference. In other words, `y` is an `&i32`, while x is an `i32`. Because every reference is a borrow, ‘`y` borrows `x`’.

    Does that answer your questions?

Leave a Reply

Would you be interested in joining LogRocket's developer community?

Join LogRocket’s Content Advisory Board. You’ll help inform the type of content we create and get access to exclusive meetups, social accreditation, and swag.

Sign up now