如何生成代码以导出Dart中节点的函数？

（我在dart2js中无法实现这一点，所以我尝试在dart开发编译器中实现；但我很乐意为dart2j找到答案！）

如果我有

test.dart

：

void activate() {
  print("activating...");
}

然后运行

dartdevc--modules节点-o test.js test.dart

输出为：

(function() {
  'use strict';
  const dart_sdk = require('dart_sdk');
  const core = dart_sdk.core;
  const dart = dart_sdk.dart;
  const dartx = dart_sdk.dartx;
  const __test = Object.create(null);
  let VoidTovoid = () => (VoidTovoid = dart.constFn(dart.definiteFunctionType(dart.void, [])))();
  __test.activate = function() {
    core.print("activating...");
  };
  dart.fn(__test.activate, VoidTovoid());
  // Exports:
  exports.__test = __test;
})();

这意味着我的函数被导出为

\u test.activate

，但我需要的是它只被

activate

我怎样才能控制这一切？JS I的目标与此相当：

exports.activate = function() { core.print("activating"); }

但我用包装纸解决了这个问题：

var extension = require('./dartvsjs/extension.js');

exports.activate = extension.__lib__extension.activate;
exports.deactivate = extension.__lib__extension.deactivate;

但我用包装纸解决了这个问题：

var extension = require('./dartvsjs/extension.js');

exports.activate = extension.__lib__extension.activate;
exports.deactivate = extension.__lib__extension.deactivate;