这段代码为何会输出father？

init();
    }

    private function init()
    {
        echo "father\n";
    }
}

class son extends father
{
    public function init()
    {
        echo "son\n";
    }
}

$son  = new son();

回复内容：

init();
    }

    private function init()
    {
        echo "father\n";
    }
}

class son extends father
{
    public function init()
    {
        echo "son\n";
    }
}

$son  = new son();

因为son里的init方法是public，而father的init方法是private，这个其实表示你son里的init方法并没有重写父类里的方法。那自然调用的仍然是父类自己的实现了

$son.init(); // son

php
class father
{
    public function __construct()
    {
        // $this->init();
      static::init();//php5.6
    }

    private function init()
    {
        echo "father\n";
    }
}

class son extends father
{
    /*public function __construct()
    {
        $this->init();
    }*/
    public function init()
    {
        echo "son\n";
    }
}

后期静态绑定

Reference: http://docs.php.com/manual/en/language.oop5.late-static-bindings.php

Note:
In non-static contexts, the called class will be the class of the object instance. Since $this-> will try to call private methods from the same scope, using static:: may give different results. Another difference is that static:: can only refer to static properties.

class father
{
    public function __construct()
    {
        $this->init();
    }

    private function init()
    {
        echo "father\n";
    }
}

class son extends father
{
    public function __construct()
    {
        parent::__construct();
        $this->init();
    }

    private function init()
    {
        echo "son\n";
    }
}

new son();

输出

father
son

建议查看__construct基础知识,想告诉楼主踏实一点,我就不信你弄清了里面的方法还会来问
授之以渔

private方法无法被重写

father中的init如果是public或protected，那么是会输出son；但是现在是private，所以在father中调用init是不会输出son的，而是调father的int输出father。