php继承
class A {public function x() {
echo "A::x() was called.\n";
}
public function y() {
self::x();
echo "A::y() was called.\n";
}
public function z() {
$this->x();
echo "A::z() was called.\n";
}
}
class B extends A {
public function x() {
echo "B::x() was called.\n";
}
}
$b = new B();
$b->y();
echo "--\n";
$b->z();
该例中，A::y()调用了A::x()，而B::x()覆盖了A::x()，那么当调用B::y()时，B::y()应该调用A::x()还是 B::x()呢？在C++中，如果A::x()未被定义为虚函数，那么B::y()（也就是A::y()）将调用A::x()，而如果A::x()使用 virtual关键字定义成虚函数，那么B::y()将调用B::x()。
然而，在PHP5中，虚函数的功能是由 self 和 $this 关键字实现的。如果父类中A::y()中使用 self::x() 的方式调用了 A::x()，那么在子类中不论A::x()是否被覆盖，A::y()调用的都是A::x()；而如果父类中A::y()使用 $this->x() 的方式调用了 A::x()，那么如果在子类中A::x()被B::x()覆盖，A::y()将会调用B::x()。
上例的运行结果如下：
A::x() was called. A::y() was called.
--
B::x() was called. A::z() was called.
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