Gulp wrap pipe

index.js

/**
 * Wrap gulp streams into fail-safe function for better error reporting
 * Usage:
 * gulp.task('less', wrapPipe(function(success, error) {
 *   return gulp.src('less/*.less')
 *      .pipe(less().on('error', error))
 *      .pipe(gulp.dest('app/css'));
 * }));
 */

function wrapPipe(taskFn) {
  return function(done) {
    var onSuccess = function() {
      done();
    };
    var onError = function(err) {
      done(err);
    }
    var outStream = taskFn(onSuccess, onError);
    if(outStream && typeof outStream.on === 'function') {
      outStream.on('end', onSuccess);
    }
  }
}

I guess you have a gulp task for mocha like this:

gulp.task('mocha', function() {
    return gulp.src('test/*.spec.js', {read: false})
        .pipe(mocha())
});

Because mocha plugin last in pipe, gulp can catch an error from here by yourself.
You don't need to add use wrapper in that case. This wrapper is made for the catching error in a middle of pipes.

Yes, I also removed wrapper from mocha task.
Thank`s for nice idea to catch build process errors!

Let's assume I have few consecutive actions in one task. I have to catch errors for each of them?
Somehow like this way?

gulp.task('styles', wrapPipe(function(success, error) {
  return gulp.src('less/*.less')
    .pipe(less().on('error', error))
    .pipe(autoprefixer().on('error', error))
    .pipe(minifyCss().on('error', error))
    .pipe(gulp.dest('app/css'));
}));

@NeXTs Yes, you should do so, because pipes do not propagate errors