How can I write a simple gulp pipe function?

Well, you don't need to use fs here, you already got the stream of file (here your chunk).

Another point, you're not sending back to the pipe the files, so I guess that's why nothing is called on your second one.

const through = require('through2')

gulp.src(sources)
  .pipe(through.obj((chunk, enc, cb) => {
    console.log('chunk', chunk.path) // this should log now
    cb(null, chunk)
  }))

In ES2015:

import through from 'through2'

gulp.src(sources)
  .pipe(through.obj((chunk, enc, cb) => cb(null, chunk)))

And for your specific example:

.pipe(through.obj((file, enc, cb) => {
  less.render(file.contents, { filename: file.path, ... }) // add other options
    .then((res) => {
      file.contents = new Buffer(res.css)
      cb(null, file)
    })
}))

This is still pretty basic, I don't check for errors, if it's not a stream and so on, but this should give you some hint on what you've missed.

How can I write a simple gulp pipe function?

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