Question

在PHP5中，我可以向类声明一个const值：

class config
{
     const mailserver = 'mx.google.com';
}

但我也可以宣布公开静态：

class config
{
     public static $mailserver = 'mx.google.com';
}

如果是配置文件我将在稍后使用，例如：

imap_connect(config::$mailserver ...
imap_connect(config::mailserver ...

您认为哪个选项更适合使用？（更快，更少的内存负载等。）

感谢。

Answer 1

静态变量可以改变，const不能改变。应该主要考虑配置变量是否应该能够在运行时更改，而不是更快。两者之间的速度差异（如果有的话）非常小，不值得考虑。

Answer 2

使用函数return global

0.0096,0.0053,0.0056,0.0054,0.0072,0.0063,0.006,0.0054,0.0054,0.0055,0.005,0.057,0.0053,0.0049,0.0064,0.054,0.0053,0.0053,0.0061,0.0059,0.0076,config1

使用get instance normal class

0.0101,0.0089,0.0105,0.0088,0.0107,0.0083,0.0094,0.0081,0.0106,0.0093,0.0098,0.0092,0.009,0.0087,0.0087,0.0093,0.0095,0.0101,0.0086,0.0088,0.0082,config2

使用static var

0.0029,0.003,0.003,0.0029,0.0029,0.0029,0.003,0.0029,0.003,0.0031,0.0032,0.0031,0.0029,0.0029,0.0029,0.0029,0.0031,0.0029,0.0029,0.0029,0.0029，config3

使用const var 0.0033,0.0031,0.0031,0.0031,0.0031,0.0031,0.0032,0.0031,0.0031,0.0031,0.0031,0.0034,0.0031,0.0031,0.0033,0.0031,0.0037,0.0031,0.0031,0.0032,0.0031，config4



function getTime() { 
    $timer = explode( ' ', microtime() ); 
    $timer = $timer[1] + $timer[0]; 
    return $timer; 
}

$config["time"] = "time";

class testconfig2
{
    public  $time = "time";
    static $instance;
    static function getInstance()
    {
        if(!isset(self::$instance))
            self::$instance = new testconfig2();
        return self::$instance;
    }
}

class testconfig3
{
    static $time = "time";
}

class testconfig4
{
    const time = "time";
}

function getConfig1()
{
    global $config;
    return $config;
}

$burncount = 2000;
$bcount = 22;

for($lcnt =1;$lcnt < $bcount;$lcnt++){
$start = getTime(); 
for($i=1;$i< $burncount;$i++)
{
    $gs=getConfig1();
    $t = $gs["time"];
} 
$end = getTime(); 
echo  round($end - $start,4).', ';
}
echo  ' config1<br/>';



for($lcnt =1;$lcnt < $bcount;$lcnt++){
$start = getTime(); 
for($i=1;$i< $burncount;$i++)
{
    $gs=testconfig2::getInstance();
    $t = $gs->time;
} 
$end = getTime(); 
echo  round($end - $start,4).', ';
}
echo  ' config2<br/>';



for($lcnt =1;$lcnt < $bcount;$lcnt++){
$start = getTime(); 
for($i=1;$i< $burncount;$i++)
{
    $gs=testconfig3::$time;
    $t = $gs;
} 
$end = getTime(); 
echo  round($end - $start,4).', ';
}
echo  ' config3<br/>';



for($lcnt =1;$lcnt < $bcount;$lcnt++){
$start = getTime(); 
for($i=1;$i< $burncount;$i++)
{
    $gs=testconfig4::time;
    $t = $gs;
} 
$end = getTime(); 
echo  round($end - $start,4).', ';
}
echo  ' config4<br/>';
?>

Answer 3

您可以将常量用于默认函数参数值，其中不允许使用静态变量。

Answer 4

const和static之间的另一个区别是类常量中不允许使用像数组这样的变量，所以

class mytest
{
    public static $c = array(1=> 'hello', 2=>'world');
}

有效，但

class mytest
{
    const c = array(1=> 'hello', 2=>'world');
}

不会。

Answer 5

另请注意，由于Yacoby上面所说的，您可以使用静态变量执行某些操作，而不能使用CONST，例如将运行时方法调用的结果分配给变量。

Answer 6

'const'和'public static'都是邪恶的！

正如Keith Humm所说 - const不能是运行时方法的结果。而且，虽然我最初认为他说你可以“将运行时方法调用的结果分配给（静态）变量”是正确的，但我的测试显示你必须跳过一些箍来做这件事。

这似乎是一个小问题，但它可能是未来灾难的种子。假设我将来被迫转向初始化方法以获得一些不可预知的原因。如果这意味着我必须在我的控制下对文件进行一些更改，我没有问题。但是，如果我将我已远程发送的库更改为其他开发人员，该怎么办？

我打算写一下，至少，如果他们想升级到我的新版本，我必须告诉其他人（以及'Future Me'）对他们的文件进行更改。在'const'的情况下，他们必须在他们使用它的地方添加'$'。在其他情况下，我今天的测试表明他们可能需要做更少但有些更改。

假设面向对象的代码，我遵循我已经看到的许多语言的一般建议，并且需要额外强调PHP - 任何类型的避免可见属性，无论是const，静态还是其他任何东西。使用'getter'方法！

这可能看起来像一个荒谬的性能打击和丑陋的代码风格，但除非你能预测未来，否则它是唯一的出路。

请参阅：Tjeerd Visser对How to initialize static variables的回答并给出了一个提示。忽略他的批评者 - 如果你关注的是性能，可能存在的错误和其他灾难，那么请切换到另一种语言。

PHP中的const vs static

6 个答案: