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)

我有点纠结:

1;

在perl中，模块需要返回true。

在Forth中，任何不包含空格的东西都可以是标识符(包含空格的东西需要做一些工作)。解析器首先检查事物是否定义了，在这种情况下，它被称为单词，如果没有，则检查它是否为数字。没有关键字。

无论如何，这意味着人们可以重新定义一个数字来表示其他东西:

: 0 1 ;

它创建了单词0，由1组成，不管它在执行时是什么。反过来，它会导致以下结果:

0 0 + .
2 Ok

另一方面，定义可以接管解析器本身——这是可以做到的 通过评论词。这意味着Forth程序实际上可以在中途变成一个完全不同语言的程序。事实上，这是Forth推荐的编程方式:首先你写你想要解决问题的语言，然后你解决问题。

让我们为所有试图废除保留词的语言(如PL/I)投票。

还有什么地方可以合法地写出这样有趣的表达:

IF IF THEN THEN = ELSE ELSE ELSE = THEN

(IF, THEN, ELSE是变量名)

or

IF IF THEN THEN ELSE ELSE

(IF为变量，THEN和ELSE为子程序)

交替:在许多语言中的事物之间交替:

boolean b = true;
for(int i = 0; i < 10; i++)
  if(b = !b)
    print i;

乍一看，b怎么可能不等于它自己呢? 这实际上只会打印奇数

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。