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。

早期的FORTRAN，空格不重要。(anti-Python !)

DO 20 I = 1, 10

含义:从这里循环到第20行，I从1到10。

DO 20 I = 1. 10

含义:将1.10分配给名为DO20I的变量。

有传言说这个漏洞毁了一个太空探测器。

JCL条件返回执行。

//STEP02 EXEC PGM=PROG02,COND=(4,GT,STEP01) .

该特性允许您根据前面步骤的返回代码运行或不运行某个步骤。真是个不错的功能。

除了一些小功能，把逻辑颠倒过来。

首先，如果条件为真，则不运行步骤。

其次，4 GT,STEP01实际上意味着“如果STEP01的返回码大于4”

因此，整个过程意味着“如果来自STEP01的返回代码大于4，则不要运行此步骤”。这与“如果4大于来自STEP01的返回代码，则运行步骤”的简单解释几乎相同，但并不完全相同。

考虑到你唯一认真看待这些事情的时间是凌晨2点30分左右，电话另一端是一个疯狂的夜班接线员，这种双重模糊会导致严重的头痛。

ActionScript 3:

当一个对象被它的接口使用时，编译器不识别从object继承的方法，因此:

IInterface interface = getInterface();
interface.toString();

给出一个编译错误。 解决方法是将类型转换为Object

Object(interface).toString();

PHP:

． 和+运算符。它有其合理的解释，但“a”+“5”= 5仍然显得尴尬。

Java(以及任何IEEE754的实现):

System.out.println(0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1);

输出0.9999999999999999

我很惊讶居然没有人提到Visual Basic的7个循环结构。

For i As Integer = 1 to 10 ... Next
While True ... End While
Do While True ... Loop
Do Until True ... Loop
Do ... Loop While True
Do ... Loop Until True
While True ... Wend

因为粘!你面前的条件实在是太复杂了!

Perl充满了奇怪但整洁的特性。

If可以用在语句之前或之后，就像这样:

print "Hello World" if $a > 1;    
if ($a > 1) { print "Hello World"; }

foreach也是如此:

print "Hello $_!\n" foreach qw(world Dolly nurse);