Antonio Scandurra (@as_

Antonio Scandurra

615 posts

Antonio Scandurra

@as__cii

Co-founder @zeddotdev. Previously: Rust and CRDTs at @dittolive, core engineer on @AtomEditor at @GitHub

Italy

Joined March 2010

Antonio Scandurra
@as__cii
Feb 5, 2024
It’s hard to describe how good Zed is feeling on this latest preview release. Typing and scrolling are just so much smoother than any other app I’ve used in my life.
Zed
@zeddotdev
Feb 5, 2024
Believe it or not, the current preview release makes Zed even faster! On ProMotion displays, we now drive your display at 120Hz during active editing. We also now apply triple buffering in our rendering pipeline for improved throughput. #CodeAtMaxFrameRate
00:00
18K
Antonio Scandurra
@as__cii
Jan 24, 2024
3 years after the first commit… we're finally open source! So excited to finally start building Zed in public!
Zed
@zeddotdev
Jan 24, 2024
Zed is now open source! Our mission is to revolutionize developer collaboration, building the world's best code editor in the process. After working together in Zed as a small team, it's time to scale up to transform Zed into a collaboration platform.
4.8K
Antonio Scandurra
@as__cii
Jun 17, 2022
Multi-buffers in @zeddotdev feel like a superpower. I was refactoring a piece of Zed's codebase and decided to record it to demonstrate how much of a time-saver it can be: it would have literally taken me twice as much using any other editor.
Antonio Scandurra
@as__cii
Apr 9, 2024
Look at us! @thorstenball and I looking like real YouTubers!
Zed
@zeddotdev
Apr 9, 2024
We started a new blog/video series: Zed Decoded. In the first episode, Thorsten & Antonio look at async Rust and how it's used in Zed.
4.3K
Antonio Scandurra
@as__cii
Mar 9, 2024
Tell me you're having fun without telling me you're having fun.
2.1K
Antonio Scandurra
@as__cii
Feb 13, 2024
I am really enjoying doing these! Part of it is because @thorstenball is such a good interviewer. But it’s also a nice way of retrospecting on the long journey that led us to where we are with Zed today.
Thorsten Ball
@thorstenball
Feb 13, 2024
Last week I sat down again with Zed's three co-founders and asked them a lot of questions: Why Rust? What problems did you run into with Atom? Why own the full tech stack? Video and transcript are live now.
3.4K
Antonio Scandurra
@as__cii
Mar 11, 2024
🥲
9.5K
Antonio Scandurra
@as__cii
Mar 9, 2024
Replying to @as__cii
(And yes, Zed can redraw a full-screen window with two editor splits in 0.8ms)
515
Antonio Scandurra
@as__cii
Jun 23, 2022
I randomly came across this paper: web.mit.edu/6.857/OldStuff… The use of Lagrangian interpolation to split data into N chunks that can be recovered using only k (< N) chunks is brilliant.
Antonio Scandurra
@as__cii
Mar 7, 2023
Excited for this blog post! It's a deep dive into some of the techniques we used to draw Zed's user interface, check it out!
Zed
@zeddotdev
Mar 7, 2023
Check out our latest blog post: "Introducing GPUI: the GPU-accelerated UI framework powering Zed" zed.dev/blog/videogame
655
Antonio Scandurra
@as__cii
Nov 10, 2016
Replying to @as__cii @giuseppegurgone and @jlongster
anyways, we are going to publish a more in-depth blog post on the Shadow DOM removal soon, so stay tuned! 😊
Antonio Scandurra
@as__cii
Nov 14, 2016
Replying to @as__cii @giuseppegurgone and @jlongster
blog.atom.io/2016/11/14/rem… I hope you enjoy reading it! :)
Antonio Scandurra
@as__cii
Feb 18, 2019
Truly remarkable paper. I loved reading it, especially the Shuffle Vector section, the graph formulation of the problem and its simplification thanks to probability theory. 💯
Gabriel Parmer (also @[email protected])
@__gparmer
Feb 17, 2019
A really cool memory allocation technique similar to compaction but for C. Mesh by @emeryberger and co. Great visual representation of the technique as well! arxiv.org/abs/1902.04738
Antonio Scandurra
@as__cii
Mar 17, 2020
Why does a complete tree of height h have at most 2^(h+1) - 1 nodes? At height 0 it has 2^0 = 1 nodes. At height 1 it has 2^1 = 2 nodes. … At height h it has 2^h nodes. So the max number of nodes is 2^0 + 2^1 + … + 2^h. Note how in base 2 this corresponds to a string of 1s.