dousao1175 2016-01-01 17:41
浏览 84
已采纳

PHP闭包和回调

I'm learning closures but I'm stuck with this:

function addPrefix($string) {
    return function($prefix) use ($string) {
        echo $prefix.$string;
    };
}
$randomstring = "a test";
$c = addPrefix($randomstring);
echo $c("This is ");

Why is $prefix concatenated? It's not even called as an argument, I just don't get it.

  • 写回答

1条回答 默认 最新

  • duandan5471 2016-01-01 17:52
    关注

    Pay attention in your example there is 2 functions. addPrefix, and an anonymous function it addPrefix returns.

    So, $c is this anonymous function (returned by addPrefix), which has the $prefix argument.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 ecplise在连接数据库时显示加载驱动成功但是数据库连接失败
  • ¥15 visionmaster启动失败,提示为“机器不满足授权而被禁用”
  • ¥50 用logisim设计16位单时钟周期cpu
  • ¥15 IDEA中圈复杂度如何具体设置
  • ¥50 labview采集不了数据
  • ¥15 Multisim红外倒车雷达仿真中距离问题
  • ¥15 请上面代码做什么处理或什么混淆
  • ¥15 英雄联盟自定义房间置顶
  • ¥15 W5500网线插上无反应
  • ¥15 如何用字典的Key,显示在WPF的xaml中