当直接向承诺传递gulp回调时出现异常

Question

我定义了“清洁代码”任务和函数“干净”任务如下

gulp.task('clean-code', function (done) {
 var files = ...;
 clean(files, done);
});

function clean (path, done) {
 del(path).then(done);
}

得到了错误

/usr/local/bin/node /usr/local/lib/node_modules/gulp/bin/gulp.js --color --gulpfile /Users/[path to project]/Gulpfile.js clean-code
[11:45:04] Using gulpfile /Users/[path to project]/Gulpfile.js
[11:45:04] Starting 'clean-code'...
[11:45:04] Cleaning: ./.tmp/**/*.js,./build/**/*.html,./build/js/**/*.js
[11:45:04] 'clean-code' errored after 8.68 ms
[11:45:04] Error
    at formatError (/usr/local/lib/node_modules/gulp/bin/gulp.js:169:10)
    at Gulp.<anonymous> (/usr/local/lib/node_modules/gulp/bin/gulp.js:195:15)
    at emitOne (events.js:77:13)
    at Gulp.emit (events.js:169:7)
    at Gulp.Orchestrator._emitTaskDone (/Users/[path to project]/node_modules/gulp/node_modules/orchestrator/index.js:264:8)
    at /Users/[path to project]/node_modules/gulp/node_modules/orchestrator/index.js:275:23
    at finish (/Users/[path to project]/node_modules/gulp/node_modules/orchestrator/lib/runTask.js:21:8)
    at cb (/Users/[path to project]/node_modules/gulp/node_modules/orchestrator/lib/runTask.js:29:3)

但是当我用下面的方式重构函数‘干净’时，一切都好了

function clean (path, done) {
  var f = function () {
    done();
  };
  del(path).then(f);
}

我不明白这有什么区别，为什么用f包装能使任务工作

Answer 1

假设您使用的是这库，那么del函数返回的承诺实际上会返回一个参数paths，这是不值得的。您可以验证是否存在这样的论点：

function clean (path, done) {
    del(path).then(function(paths) {
        console.log(paths);
        done();
    });
}

在您的代码中：

function clean (path, done) {
    del(path).then(done);
}

将paths参数转发给done函数，后者将将其解释为导致应用程序崩溃的错误参数。通过自己调用done()，不会转发任何参数，任务将被正确执行。