如何评估条件中的所有表达式？

Question

我知道PHP使用懒惰评估/短路运营商。但说我想要评估所有表达式的条件，例如：

$a = "Apple"; 
$b = "Banana"; 
$c = "Cherry"; 

function check($fruit) { 
    if ($fruit != "Banana") { 
     echo "$fruit is good.\n"; 
     return true; 
    } else { 
     echo "$fruit is bad.\n"; 
     return false; 
    } 
} 

if (check($a) && check($b) && check($c)) { 
    echo "Yummy!\n"; 
} 

因为懒惰的评价，这样只会输出：

Apple is good. 
Banana is bad. 

比期望的输出，而：

Apple is good. 
Banana is bad. 
Cherry is good. 

这在表单验证中很有用。

所以我的问题：有没有什么办法来强制所有的表达式在PHP中进行评估，如果不是，在上面的例子中最好/最快的方式来获得所需的结果？

Answer 1

您可以使用bitwise AND（单个号：&）

$a = "Apple"; 
$b = "Banana"; 
$c = "Cherry"; 

function check($fruit) { 
    echo ($fruit != "Banana") ? "$fruit is good.\n" : "$fruit is bad.\n"; 
} 

if (check($a) & check($b) & check($c)) { 
    echo "Yummy!\n"; 
} 

打印：

苹果是不错的。

香蕉不好。

樱桃很好。

例子：http://sandbox.onlinephpfunctions.com/code/07092a9d6636ae8ddafce024d7cc74643e311e9c

Answer 2

function check($fruit) { 
    echo ($fruit != "Banana") ? "$fruit is good.\n" : "$fruit is bad.\n"; 
    return $fruit != "Banana"; 
} 


$a = "Apple"; 
$b = "Banana"; 
$c = "Cherry"; 
if (check($a) & check($b) & check($c)) { 
    echo "Yummy!\n"; 
} 


/* 
Apple is good. 
Banana is bad. 
Cherry is good. 
*/ 


$a = "Apple"; 
$b = "apple"; 
$c = "Cherry"; 
if (check($a) & check($b) & check($c)) { 
    echo "Yummy!\n"; 
} 


/* 
Apple is good. 
apple is good. 
Cherry is good. 
Yummy! 
*/