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 to solve coordination problems in Islands architecture

How to solve coordination problems in Islands architecture

Solve coordination problems in Islands architecture using event-driven patterns instead of localStorage polling.

Muhammed Ali
Feb 26, 2026 ⋅ 5 min read
lewis angular signal forms

Signal Forms: Angular’s best quality of life update in years

Signal Forms in Angular 21 replace FormGroup pain and ControlValueAccessor complexity with a cleaner, reactive model built on signals.

Lewis Cianci
Feb 25, 2026 ⋅ 10 min read
replay 2 25 26

The Replay (2/25/26): Signal Forms, Ralph to the rescue, and more

Discover what’s new in The Replay, LogRocket’s newsletter for dev and engineering leaders, in the February 25th issue.

Matt MacCormack
Feb 25, 2026 ⋅ 32 sec read

Google & Shopify’s UCP: How AI agents sell online

Explore how the Universal Commerce Protocol (UCP) allows AI agents to connect with merchants, handle checkout sessions, and securely process payments in real-world e-commerce flows.

Emmanuel John
Feb 24, 2026 ⋅ 8 min 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

Hey there, want to help make our blog better?

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