大约20年前，我用一个编译器为一种叫做Coral的语言工作，它允许我声明只写变量!

不过，这是有道理的，因为它们是全球性的，被用作一种信号机制。一个进程写入值，另一个进程读取值。

在Python中，函数参数的“编译时”(或声明时)计算可能令人困惑:

def append(v, l = []):
    l.append(v)
    return l


print append(1)
print append(2)

>>> [1]
>>> [1,2]

其意图可能是:

def append(v, l = None):
    if l is None:
        l = []
    l.append(v)
    return l

print append(1)
print append(2)

>>> [1]
>>> [2]

这种行为对于缓存之类的事情很有用，但它可能是危险的。

附加特性:具有可变内容的元组:

a = (1,2,[3])
a[2][:] = [4] # OK
a[2] = [2] # crashes

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)

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。

为什么c#的List<T>.AddRange()不让我添加T的子类型的元素?阀门列表< T > () ! 微软只需要多写一行代码:

public void AddRange<S>(
    IEnumerable<S> collection
) where S : T

好吧，既然问题是断断续续的，我就加入到“有趣”中来

Go(又名Issue9)使用大写的可见性:

如果你用大写字母命名某个东西，它就会被公众访问。 如果你使用小写，它将被包保护:

包装外可见:

func Print(v ...) { 
}

包外不可见

func print( v ... ) {
}

你可以在这个原始答案中找到更多。