如何检查派生类中的函数是否被覆盖？

Question

我有一个带有函数的基类，还有几个从基类派生的类。这些派生类中有一些重写该功能，有些则没有。

是否有一种方法可以检查某个特定的对象（该对象是已知的派生类之一）是否已覆盖该函数？

示例：

<?php

class BaseThing
{
    function Bla() { echo "Hello, this is the base class\n"; }
}

class DerivedThing extends BaseThing
{
    function Bla() { echo "Hello, this is a derived class\n"; }
}

class AnotherDerivedThing extends BaseThing
{
    // Does not override Bla()
}

$a = new BaseThing();
$b = new DerivedThing();
$c = new AnotherDerivedThing();

$a->Bla(); // prints base class
$b->Bla(); // prints derived class
$c->Bla(); // prints base class

if (method_exists($b,'Bla')) echo "Method 'Bla' exists in DerivedThing\n";
if (method_exists($c,'Bla')) echo "Method 'Bla' exists in AnotherDerivedThing\n";

?>

[我尝试使用method_exists，但显然它说$c包含该方法，因为它是从包含该方法的类派生的。

是否可以检查对象是否覆盖特定功能？例如。在上面的示例中，我可以以某种方式检测到$b确实覆盖了Bla()函数，但是$c没有吗？

Answer 1

您可以使用ReflectionClass::getMethod()并比较方法：

<?php
class BaseThing
{
    function Bla() { echo "Hello, this is the base class\n"; }
}

class DerivedThing extends BaseThing
{
    function Bla() { echo "Hello, this is a derived class\n"; }
}

class AnotherDerivedThing extends BaseThing
{
    // Does not override Bla()
}

$reflectorBase = new ReflectionClass('BaseThing');
$reflectorDerived = new ReflectionClass('DerivedThing');
$reflectorAnotherDerived = new ReflectionClass('AnotherDerivedThing');

if ($reflectorBase->getMethod('Bla') == $reflectorDerived->getMethod('Bla'))
{
    echo "methods are same in base and derived" . PHP_EOL;
}
else
{
    echo "methods are NOT same in base and derived" . PHP_EOL;
}

if ($reflectorBase->getMethod('Bla') == $reflectorAnotherDerived->getMethod('Bla'))
{
    echo "methods are same in base and derived 2" . PHP_EOL;
}
else
{
    echo "methods are NOT same in base and derived 2" . PHP_EOL;
}

此输出：

methods are NOT same in base and derived
methods are same in base and derived 2