Coffeescript 如何将参数传递到一个gulp任务回调中？

我试着做两个任务，一个监视和构建任务。 watch任务调用我的'coffee'任务，它将我的

.coffee

文件编译成javascript。 构建任务基本上也应该这样做，只是我想将布尔值解析到函数中，这样我就可以编译包含源映射的代码

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']

有人能解决我的问题吗？我原则上做错了什么吗？

---

提前感谢。

咖啡任务不一定是一项狼吞虎咽的任务。让它成为一个JavaScript函数

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']

---

我想你指的是“传递”参数，而不是“解析”参数？当然谢谢@jbyrd