在PHP中，你可以这样做:

System.out.print("hello");

Python的everything-is-really-a-reference有一个有趣的副作用:

>>> a = [[1]] * 7
>>> a
[[1], [1], [1], [1], [1], [1], [1]]
>>> a[0][0] = 2
>>> a
[[2], [2], [2], [2], [2], [2], [2]]

在JavaScript中，void不是关键字，不是类型声明，也不是变量名，也不是函数，也不是对象。Void是一个前缀操作符，类似于-、——、++和!你可以给任何表达式加上前缀，这个表达式的值将是undefined。

它经常被用在bookmarklet和内联事件处理程序中，比如下面这个比较常见的例子:

<a href="javascript:void(0)">do nothing</a>

在这个例子中使用它的方式使它看起来像一个函数调用，而实际上它只是一种获得原始未定义值的过于聪明的方法。大多数人并没有真正理解JavaScript中void的真正本质，这可能会导致许多讨厌的bug和奇怪的意想不到的事情发生。

不幸的是，我认为void操作符是在JavaScript中获得未定义值的唯一真正保证的方法，因为未定义，正如在另一个回答中指出的，是一个可以重新赋值的变量名，而{}。a可以被Object.prototype.a = 'foo'打乱

更新:我想到了另一种生成undefined的方法:

(function(){}())

嗯，有点啰嗦，返回“undefined”是它的目的就更不清楚了。

PHP

PHP对实例变量和方法的重载处理不一致。考虑:

class Foo
{
    private $var = 'avalue';

    private function doStuff()
    {
        return "Stuff";
    }

    public function __get($var)
    {
        return $this->$var;
    }

    public function __call($func, array $args = array())
    {
        return call_user_func_array(array($this, $func), $args);
    }
}

$foo = new Foo;
var_dump($foo->var);
var_dump($foo->doStuff());

转储$var是有效的。即使$var是私有的，__get()也会被任何不存在或不可访问的成员调用，并返回正确的值。这不是doStuff()的情况，它失败于:

Fatal error: Call to private method Foo::doStuff() from context ”.”

我认为其中很多都是在c风格的语言中工作的，但我不确定。

Pass a here document as a function argument: function foo($message) { echo $message . "\n"; } foo(<<<EOF Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nunc blandit sem eleifend libero rhoncus iaculis. Nullam eget nisi at purus vestibulum tristique eu sit amet lorem. EOF ); You can assign a variable in an argument list. foo($message = "Hello"); echo $message; This works because an assignment is an expression which returns the assigned value. It’s the cause of one of the most common C-style bugs, performing an assignment instead of a comparison.

Python

在Python中，可变的默认函数参数会导致意想不到的结果:

def append(thing, collection=[]):
    collection.append(thing)
    return collection

print append("foo")
# -> ['foo']
print append("bar")
# -> ['foo', 'bar']
print append("baz", [])
# -> ['baz']
print append("quux")
# -> ['foo', 'bar', 'quux']

空列表是在函数定义时初始化的，而不是在调用时初始化的，因此对它的任何更改都会在函数调用之间保持不变。

MySQL的大小写敏感性

MySQL有非常不寻常的区分大小写的规则:表区分大小写，列名和字符串值不区分大小写:

mysql> CREATE TEMPORARY TABLE Foo (name varchar(128) NOT NULL);
DESCRIBE foo;
ERROR 1146 (42S02): Table 'foo' doesn't exist
mysql> DESCRIBE Foo;
+-------+--------------+------+-----+---------+-------+
| Field | Type         | Null | Key | Default | Extra |
+-------+--------------+------+-----+---------+-------+
| name  | varchar(128) | NO   |     | NULL    |       |
+-------+--------------+------+-----+---------+-------+
1 row in set (0.06 sec)
mysql> INSERT INTO Foo (`name`) VALUES ('bar'), ('baz');
Query OK, 2 row affected (0.05 sec)

mysql> SELECT * FROM Foo WHERE name = 'BAR';
+------+
| name |
+------+
| bar  |
+------+
1 row in set (0.12 sec)

mysql> SELECT * FROM Foo WHERE name = 'bAr';
+------+
| name |
+------+
| bar  |
+------+
1 row in set (0.05 sec)

在SQL

NULL不等于NULL

所以你不能:

WHERE myValue == NULL

这将总是返回false。

NULL != NULL

Haskell's use of Maybe and Just. Maybe a is a type constructor that returns a type of Just a, but Maybe Int won't accept just an Int, it requires it to be a Just Int or Nothing. So in essence in haskell parlance Just Int is about as much of an Int as an apple is an orange. The only connection is that Just 5 returns a type of Maybe Interger, which can be constructed with the function Just and an Integer argument. This makes sense but is about as hard to explain as it can theoretically be, which is the purpose of haskell right? So is Just really JustKindaLikeButNotAtAll yea sorta, and is Maybe really a KindaLooksLikeOrIsNothing, yea sorta again.

-- Create a function that returns a Maybe Int, and return a 5, which know is definitly Int'able
>  let x :: Maybe Int; x = 5;
<interactive>:1:24:
    No instance for (Num (Maybe Int))
      arising from the literal `5' at <interactive>:1:24
    Possible fix: add an instance declaration for (Num (Maybe Int))
    In the expression: 5
    In the definition of `x': x = 5

>  Just 5  
Just 5
it :: Maybe Integer

    -- Create a function x which takes an Int
>  let x :: Int -> Int; x _ = 0;
x :: Int -> Int
-- Try to give it a Just Int
>  x $ Just 5                   

<interactive>:1:4:
    Couldn't match expected type `Int' against inferred type `Maybe t'
    In the second argument of `($)', namely `Just 5'
    In the expression: x $ Just 5
    In the definition of `it': it = x $ Just 5

祝你好运读到这篇文章，我希望它是正确的。