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设计模式 - 在看深入php第8章设计原则遇到一些困惑

WBOY
WBOY원래의
2016-06-06 20:31:01939검색

下面代码是书上例子,在php下运行也没错误,但有几点不解
1.abstract 抽象类 里面竟然没有抽象方法?
2. function __construct( $duration, CostStrategy $strategy ) {
还有这种传递方式?我仔细看了一下,应该是把下面声明的抽象类CostStrategy以别名方式传进来,是这样吗,如果是真的,抽象类CostStrategy都不需要被继承吗?
这段代码把我搞糊涂了

<code><?php abstract class Lesson {
    private   $duration;
    private   $costStrategy;

    function __construct( $duration, CostStrategy $strategy ) {
        $this->duration = $duration;
        $this->costStrategy = $strategy;
    }

    function cost() {
        return $this->costStrategy->cost( $this );
    }

    function chargeType() {
        return $this->costStrategy->chargeType( );
    }

    function getDuration() {
        return $this->duration;
    }

    // more lesson methods...
}


abstract class CostStrategy {
    abstract function cost( Lesson $lesson );
    abstract function chargeType();
}

class TimedCostStrategy extends CostStrategy {
    function cost( Lesson $lesson ) {
        return ( $lesson->getDuration() * 5 );
    }
    function chargeType() {
        return "hourly rate";
    }
}

class FixedCostStrategy extends CostStrategy {
    function cost( Lesson $lesson ) {
        return 30;
    }

    function chargeType() {
        return "fixed rate";
    }
}

class Lecture extends Lesson {
    // Lecture-specific implementations ...
}

class Seminar extends Lesson {
    // Seminar-specific implementations ...
}

$lessons[] = new Seminar( 4, new TimedCostStrategy() );
$lessons[] = new Lecture( 4, new FixedCostStrategy() );

foreach ( $lessons as $lesson ) {
    print "lesson charge {$lesson->cost()}. ";
    print "Charge type: {$lesson->chargeType()}\n";
}
?>

</code>

回复内容:

下面代码是书上例子,在php下运行也没错误,但有几点不解
1.abstract 抽象类 里面竟然没有抽象方法?
2. function __construct( $duration, CostStrategy $strategy ) {
还有这种传递方式?我仔细看了一下,应该是把下面声明的抽象类CostStrategy以别名方式传进来,是这样吗,如果是真的,抽象类CostStrategy都不需要被继承吗?
这段代码把我搞糊涂了

<code><?php abstract class Lesson {
    private   $duration;
    private   $costStrategy;

    function __construct( $duration, CostStrategy $strategy ) {
        $this->duration = $duration;
        $this->costStrategy = $strategy;
    }

    function cost() {
        return $this->costStrategy->cost( $this );
    }

    function chargeType() {
        return $this->costStrategy->chargeType( );
    }

    function getDuration() {
        return $this->duration;
    }

    // more lesson methods...
}


abstract class CostStrategy {
    abstract function cost( Lesson $lesson );
    abstract function chargeType();
}

class TimedCostStrategy extends CostStrategy {
    function cost( Lesson $lesson ) {
        return ( $lesson->getDuration() * 5 );
    }
    function chargeType() {
        return "hourly rate";
    }
}

class FixedCostStrategy extends CostStrategy {
    function cost( Lesson $lesson ) {
        return 30;
    }

    function chargeType() {
        return "fixed rate";
    }
}

class Lecture extends Lesson {
    // Lecture-specific implementations ...
}

class Seminar extends Lesson {
    // Seminar-specific implementations ...
}

$lessons[] = new Seminar( 4, new TimedCostStrategy() );
$lessons[] = new Lecture( 4, new FixedCostStrategy() );

foreach ( $lessons as $lesson ) {
    print "lesson charge {$lesson->cost()}. ";
    print "Charge type: {$lesson->chargeType()}\n";
}
?>

</code>

function __construct( $duration, CostStrategy $strategy )
这行代码中的CostStrategy $strategy是指定了该参数的类型约束,必须是CostStrategy类或者其子类的对象

PHP类型约束:http://php.net/manual/zh/language.oop5.typehinting.php
至于抽象类里没有抽象方法,确实是可以这样的,官方文档里的解释要细细读一下比较绕
PHP 5 支持抽象类和抽象方法。定义为抽象的类不能被实例化。任何一个类,如果它里面至少有一个方法是被声明为抽象的,那么这个类就必须被声明为抽象的
就是说如果有抽象方法了那么这个类必须声明成抽象类,抽象方法是抽象类的必要条件,而抽象类不是抽象方法的充分条件, 不知这么解释是否清楚。
PHP 抽象类 : http://php.net/manual/zh/language.oop5.abstract.php

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