How to throw an error when executing Gulp without stopping it?

Question

gulp.js

Here is gulpfile.js

var gulp = require('gulp'),
  less = require('gulp-less');

gulp.task('less', function() {
  gulp.src('css/style.less')
    .pipe(less())
    .pipe(gulp.dest('css/'));
});

gulp.task('watch', function() {
  gulp.watch('css/*.less', ['less']);
})

gulp.task('default', ['watch']);

If I make a mistake in the syntax, then an error is displayed in the console and gulp stops.
Is it possible to make it so that it would show an error notification in a separate window and not have to run gulp again in the console.

Answer 1

i use gulp-plumber it catches errors and prints them to the terminal without stopping the task.

var gulp = require('gulp'),
  less = require('gulp-less');
var plumber = require('gulp-plumber');

gulp.task('less', function() {
  gulp.src('css/style.less')
    .pipe(plumber())
    .pipe(less())
    .pipe(gulp.dest('css/'));
});

Answer 2

Thanks