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:

the replay nov 19

The Replay (11/19/25): React 19.2: The async shift is finally here

Discover what’s new in The Replay, LogRocket’s newsletter for dev and engineering leaders, in the November 19th issue.

Matt MacCormack
Nov 19, 2025 ⋅ 33 sec read

React 19.2: The async shift is finally here

Jack Herrington writes about how React 19.2 rebuilds async handling from the ground up with use(), , useTransition(), and now View Transitions.

Jack Herrington
Nov 19, 2025 ⋅ 5 min read

Offline-first frontend apps in 2025: IndexedDB and SQLite in the browser and beyond

The web has always had an uneasy relationship with connectivity. Most applications are designed as if the network will be […]

Alexander Godwin
Nov 18, 2025 ⋅ 11 min read
Real-Time AI In Next.js How To Stream Responses With The Vercel AI SDK

Real-time AI in Next.js: How to stream responses with the Vercel AI SDK

Streaming AI responses is one of the easiest ways to improve UX. Here’s how to implement it in a Next.js app using the Vercel AI SDK—typing effect, reasoning, and all.

Elijah Asaolu
Nov 17, 2025 ⋅ 9 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

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