Gulp swallowing eslint output if there are too many files

Question

Consider the following gulp file:

var gulp = require('gulp');
var eslint = require('gulp-eslint');
var debug = require('gulp-debug');

gulp.task('lint', function (done) {
    gulp
        .src([
            'src/**/*.js',
            '!src/public/javascripts/external/*'
        ])
        .pipe(eslint())
        .pipe(eslint.format())
        //.pipe(debug({title: 'Linting:'}));
    done();
});

If my src folder contains too many files (I am not talking about an excessive number. It's less than 20), then gulp lint will only output

Using gulpfile [my/path/to/gulpfile]
Starting 'lint'...
Finished 'lint' after 55ms

There won't be any warnings from ESLint, even though I made sure there are problems in my code of course. This problem can be reproduced by manually adding javascript files from my src folder without using wildcards. After a certain number of files (I sadly forgot to count), errors won't be displayed any more. This does depend not on which files I add, just the number.

For some reason this behavior can be 'fixed' by adding the commented line that outputs debug information, so I am assuming my mistake has something to do with me misunderstanding how the gulp works internally. ESLint also works fine when called externally. Any ideas what the problems could be or steps to narrow it down?

Answer 1

I was able to fix my problem although I am not 100% sure what the problem was. According to the gulp-eslint package description you are supposed to return the result of the pipes. So the correct gulpfile would look like this:

var gulp = require('gulp');
var eslint = require('gulp-eslint');
var debug = require('gulp-debug');

gulp.task('lint', function () {
    return gulp // note the return here
        .src([
            'src/**/*.js',
            '!src/public/javascripts/external/*'
        ])
        .pipe(eslint())
        .pipe(eslint.format());
    // no call to 'done()' is needed
});

My guess is that the plugin runs asynchronously and I ended the task by calling done() before it was actually done. Printing the debug information either happened after the asynchronous task was done or it bought enough time to finish. Now gulp will properly receives a promise (or something like that) and waits until it is finished.

Can anyone confirm this guess?