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关于PHP中函数的返回值的有关问题

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Release: 2016-06-13 13:30:18
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关于PHP中函数的返回值的问题

PHP code
<!--

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-->function Traverse_files($dir1,$dir2){
  $dir = $dir1.'/'.$dir2;
  //print $dir;
  //exit;
  $ss = array();
  if (is_dir($dir)){
    if ($dh = opendir($dir)){
      while (($file = readdir($dh)) !== false){
        if($file!=="."&& $file!==".."){
          $ss[]=$file;
        }
      }
      print_r($ss);
      return $ss;
      closedir($dh);
    }
  }
}
$dirbase = 'd:';
$dirito = '5-24';
$bb = Traverse_files($dirbase,$dirito);
printf($bb);

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上代码,问题是上面retrun的值出不来。。$ss在上面可以打印出来。
如何解决。。

------解决方案--------------------
printf() 函数输出格式化的字符串。 应该用 print_r($bb);
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