红宝石拖鞋。条件语句中的"…"和".."并不总是范围操作符:

(0..20).each do |x|
  if ((x%10) == 5)..((x%10) == 5)
    print "#{x} "
  end
end

(0..20).each do |x|
  if ((x%10) == 5)...((x%10) == 5)
    print "#{x} "
  end
end

这将输出:

5 15
5 6 7 8 9 10 11 12 13 14 15

. .检查每一遍的两个语句，…每次只检查“on”或“off”语句(取决于触发器状态)。它们是从awk和sed偷来的。

Matz在“Ruby编程语言”中写道:“人字拖是Ruby中一个相当晦涩的特性，可能最好避免……”

C的多个名称空间:

typedef int i;

void foo()
{
    struct i {i i;} i;
    i: i.i = 3;
    printf( "%i\n", i.i);
}

或与字符:

typedef char c;

void foo()
{
    struct c {c c;} c;
    c: c.c = 'c';
    printf( "%c\n", c.c);
}

在javaScript中，NaN是一个全局变量。

foo是什么数据类型?

SELECT TOP 1
   NULL AS foo
INTO
   dbo.bar
FROM
   sys.columns --trivial

为什么所有东西都归零?

SELECT CAST('' AS int), CAST('' AS datetime), CAST('' AS float)

.．.除了这

SELECT CAST('' AS decimal)

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。

Perl中的字符串数学非常奇怪。

$ perl -E '$string = "a"; $string++; say $string'
b

$ perl -E '$string = "abc"; $string++; say $string'
abd

$ perl -E '$string = "money"; $string++; say $string'
monez

$ perl -E '$string = "money"; $string--; say $string'
-1