How can I pass arguments into a gulp task callback? - coffeescript

Question

I am trying to make two tasks, a watch and build task.
The watch task calls my 'coffee' task wich compiles my .coffee files into javascript.
The build task should basically do the same, except that I wanna parse a boolean into the function, so that I can compile the code including source maps.
gulp = require 'gulp'
gutil = require 'gulp-util'
clean = require 'gulp-clean'
coffee = require 'gulp-coffee'
gulp.task 'clean', ->
gulp.src('./lib/*', read: false)
.pipe clean()
gulp.task 'coffee', (map) ->
gutil.log('sourceMap', map)
gulp.src('./src/*.coffee')
.pipe coffee({sourceMap: map}).on('error', gutil.log)
.pipe gulp.dest('./lib/')
# build app
gulp.task 'watch', ->
gulp.watch './src/*.coffee', ['coffee']
# build app
gulp.task 'build', ->
gulp.tasks.clean.fn()
gulp.tasks.coffee.fn(true)
# The default task (called when you run `gulp` from cli)
gulp.task 'default', ['clean', 'coffee', 'watch']
Does somebody have a solution for my problem? Am I doing something in principle wrong?
Thanks in advance.

Answer 1

The coffee task need not be a gulp task. Just make it a JavaScript function.
gulp = require 'gulp'
gutil = require 'gulp-util'
clean = require 'gulp-clean'
coffee = require 'gulp-coffee'
gulp.task 'clean', ->
gulp.src('./lib/*', read: false)
.pipe clean()
compile = (map) ->
gutil.log('sourceMap', map)
gulp.src('./src/*.coffee')
.pipe coffee({sourceMap: map}).on('error', gutil.log)
.pipe gulp.dest('./lib/')
# build app
gulp.task 'watch', ->
gulp.watch './src/*.coffee', =>
compile(false)
# build app
gulp.task 'build', ['clean'], ->
compile(true)
# The default task (called when you run `gulp` from cli)
gulp.task 'default', ['clean', 'build', 'watch']