輸出是不希望

Question

我試圖讓基本這兒類的定義和使用計算

下面的代碼

<?php 

class calculator { 
    var $number1 = 4; 
    var $number2 = 5; 

    function add ($a,$b){ 
     $c = $a + $b; 
     print ("the sum of your numbers: $c"); 
     print ($c); 
    } 

} 

$cal = new calculator; 
$cal->add($number1,$number2); 

?> 

什麼出現在我的瀏覽器：

您的號碼之和：0

爲什麼不是9？

Answer 1

您傳遞的$number1和$number2的值是多少？ $number1和$number2與$cal->number1和$cal->number2不相同。

您正在定義對象的兩個屬性，並將兩個不同的單獨變量傳遞到類的函數中。基本上有兩對數字 - 一個在對象中，數值爲4和5，一個在函數的外部，沒有任何值（都是0），然後再添加。

你可以試試這個：

<?php 
class calculator { 
    private $number1 = 4; 
    private $number2 = 5; 

    function add ($a, $b){ 
     $c = $this->$a + $this->$b; 
     print ("the sum of your numbers: $c"); 
     print ($c); 
    } 

} 

$cal = new calculator; 
$cal->add('number1', 'number2'); 

或者這樣：

<?php 
class calculator { 
    private $number1 = 4; 
    private $number2 = 5; 

    function add(){ 
     $c = $this->number1 + $this->number2; 
     print ("the sum of your numbers: $c"); 
     print ($c); 
    } 

} 

$cal = new calculator; 
$cal->add(); 

或者這樣：

<?php 
class calculator { 
    function add ($a, $b){ 
     $c = $a + $b; 
     print ("the sum of your numbers: $c"); 
     print ($c); 
    } 

} 

$cal = new calculator; 
$cal->add(4, 5); 

Answer 2

你$number1和$number2正在類的範圍內聲明。

但是，當您撥打$cal->add($number1, $number2)時，您現在不在該範圍之內，因此這些值未定義。

Answer 3

您應該做

class calculator { 
    //... 
} 
$number1 = 4; 
$number2 = 5; 
$cal = new calculator; 
$cal->add($number1,$number2); 

或

class calculator { 
    var $number1 =4; 
    var $number2 =5; 
    //... 
} 
$cal = new calculator; 
$cal->add($cal->number1,$cal->number2);