2020-08-27
1731
#rust
Thomas Heartman
23920
Aug 27, 2020 â‹… 6 min read

Understanding lifetimes in Rust

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

Recent posts:

When is low-code the right choice? Here’s how to decide

Not sure if low-code is right for your next project? This guide breaks down when to use it, when to avoid it, and how to make the right call.

Popoola Temitope
Jul 11, 2025 â‹… 7 min read
Comparing AI App Builders — Firebase Studio vs. Lovable vs. Replit. LogRocket Article

Comparing AI app builders — Firebase Studio vs. Lovable vs. Replit

Compare Firebase Studio, Lovable, and Replit for AI-powered app building. Find the best tool for your project needs.

Emmanuel John
Jul 11, 2025 â‹… 7 min read
Gemini CLI tutorial — Will it replace Windsurf and Cursor?

Gemini CLI tutorial — Will it replace Windsurf and Cursor?

Discover how to use Gemini CLI, Google’s new open-source AI agent that brings Gemini directly to your terminal.

Chizaram Ken
Jul 10, 2025 â‹… 8 min read
React & TypeScript: 10 Patterns For Writing Better Code

React & TypeScript: 10 patterns for writing better code

This article explores several proven patterns for writing safer, cleaner, and more readable code in React and TypeScript.

Peter Aideloje
Jul 10, 2025 â‹… 11 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