title: How to resolve "Using a domain property in MakeCallback is deprecated" warning when running gulp tasks?
body: <p>I'm using node with gulp to run some building tasks. This worked fine until a couple days ago. Now (I assume after an upgrade/update, not sure which specific one. I believe it's the update of node from 14.4 to 14.5) I keep getting this warning</p>
  <pre><code>[DEP0097] DeprecationWarning: Using a domain property in MakeCallback is deprecated. Use the async_context variant of MakeCallback or the AsyncResource class instead.
  (Use `node --trace-deprecation ...` to show where the warning was created)
  </code></pre>
  <p>I cannot figure out how to use <code>--trace-deprecation</code> with gulp so I couldn't find what is triggering it.<br />
  My actual <em>gulpfile</em> is much longer and commenting out sections, changing <code>pipeline</code> to <code>.pipe</code>, updating node and dependencies, using <code>async/await</code>, and some other minor changing didn't get me any closer to narrowing down what the issue is.</p>
  <p>Therefore I set up this minimal working example below:</p>
  <ul>
  <li>running the default gulp task (<code>clean_fake</code>) does not trigger the warning</li>
  <li>both <code>gulp clean</code> and <code>gulp styles</code> cause the warning to show</li>
  </ul>
  <pre class="lang-js prettyprint-override"><code>const gulp = require('gulp');
  const del = require('del');
  const sass = require('gulp-sass');
  
  async function clean() {
      const deletedPaths = await del([ './js/*.site.js', './style.css' ], { dryRun: true });
      console.log('Deleted files and directories:\n', deletedPaths.join('\n'));
  }
  
  async function clean_fake() {
      const deletedPaths = await test();
      console.log('Deleted files and directories:\n', deletedPaths.join('\n'));
  }
  function test() {
      console.log('dummy function');
      return [ 'test' ];
  }
  
  function styles() {
      return gulp.src('./src/sass/style.scss').pipe(sass()).pipe(gulp.dest('./'));
  }
  
  exports.clean = clean;
  exports.styles = styles;
  
  exports.default = clean_fake;
  </code></pre>
  <p><strong>Current versions:</strong><br />
  node: v14.5.0<br />
  npm: 6.14.6</p>
  <p>del: 5.1.0<br />
  gulp: 4.0.2<br />
  gulp-sass: 4.1.0</p>
  <p>PS: There is this similar <a href="https://stackoverflow.com/questions/56440304/nodejs-dep0097-deprecationwarning">question</a>, but with no answer for my problem.</p>
  <h2>Update:</h2>
  <p>I figured out how to trace deprecation by running it with <code>NODE_OPTIONS</code>:</p>
  <pre><code>NODE_OPTIONS='--trace-deprecation' gulp
  </code></pre>
  <p>However the output didn't help me much</p>
  <pre><code>(node:146806) [DEP0097] DeprecationWarning: Using a domain property in MakeCallback is deprecated. Use the async_context variant of MakeCallback or the AsyncResource class instead.
      at emitMakeCallbackDeprecation (domain.js:123:13)
      at FSReqCallback.topLevelDomainCallback (domain.js:134:5)
      at FSReqCallback.callbackTrampoline (internal/async_hooks.js:121:14)
  </code></pre>