title: Gulp: Use output from one task in another task
body: <p>I have have three gulp tasks, where the last task (<code>allScripts</code>) runs the two dependent tasks first and then joins the resultant files from them. </p>
  
  <p>I could, in the last task, delete the two result-files from the two first tasks and live happily ever after with the joined file. </p>
  
  <p>But I was thinking, is it possible to avoid the two temporary files by somehow piping them into the <code>allScripts</code> task "directly"?</p>
  
  <pre><code>gulp.task('firstGroup', function() {
    return gulp.src('some/files/*.js')
      .pipe(doSomething())
      .pipe(concat('some-scripts.js'))
      .pipe(gulp.dest('dest'));
  });
  
  gulp.task('secondGroup', function() {
    return gulp.src('some/other/files/*.js')
      .pipe(doSomethingElse())
      .pipe(concat('some-other-scripts.js'))
      .pipe(gulp.dest('dest'));
  });
  
  gulp.task('allScripts', ['firstGroup','secondGroup'], function() {
    return gulp.src(['dest/some-scripts.js','dest/some-other-scripts.js'])
      .pipe(concat('all-scripts.js'))
      .pipe(gulp.dest('dest'))
      // delete the two src-files
  });
  </code></pre>