How does libuv handle over four concurrent file read tasks in Node.js?

I understand that Node.js uses the libuv library for asynchronous I/O, which by default utilizes four worker threads. My question is: When more than four file read tasks are scheduled, does a libuv thread read a file entirely before switching to another, or does it interleave reading multiple files?

Below is the code that logs "data events" and "end events" from multiple read streams:

JAVASCRIPTCopied
const fs = require('node:fs');
const path = require('node:path');

function getFilePaths(directory){
  let files = fs.readdirSync(directory)
  return files
  .map((file) => path.join(directory, file))
  .filter((filePath) => fs.statSync(filePath).isFile())
}

const directory = 'D:\test-folder';

const files = getFilePaths(directory)

const streams = []

files.forEach((file, index)=>{
  streams.push([index, fs.createReadStream(file)])
})

streams.forEach(([index, stream])=>{
  stream.on('data', () => {
    console.log(`Data: Stream ${index}`);
  });

  stream.on('end', () => {
    console.log(`End: Stream ${index}`);
  });

  stream.on('error', (err) => {
    console.error(`Error: Stream ${index}`);
  });
})

In my output, I expected that at least one of the first four files would complete reading before any data chunk from the remaining files is processed. Instead, it appears that libuv threads interleave I/O operations across multiple files. Is this interleaving behavior expected?