Delphi不关心像“word”这样的类型转换，将读取数组arr[0..65535] where pos = 65535: Arr[单词(pos + 10)]

c++(或Java)中没有封装的事实。任何对象都可以违反任何其他对象的封装，破坏它的私有数据，只要它是相同的类型。例如:

#include <iostream>
using namespace std;

class X
{
  public:
    // Construct by passing internal value
    X (int i) : i (i) {}

    // This breaks encapsulation
    void violate (X & other)
    {
        other.i += i;
    }

    int get () { return i; }

  private:
    int i;
};

int main (int ac, char * av[])
{
    X a(1), b(2), c(3);

    a.violate (c);
    b.violate (c);
    cout << c.get() << endl;    // "6"
}

Java有一整本关于它们的书。

书http://www.javapuzzlers.com/lg-puzzlers-cropped.jpg

爪哇益智游戏

这并不是说它被大量使用，而是c++的“返回对静态大小数组的引用”的语法很奇怪:

struct SuperFoo {
  int (&getFoo() const)[10] {
    static int foo[10];
    return foo;
  }
}

在上述情况下，Ofc方法可以声明为静态const

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。

在Common Lisp中，零维数组是很奇怪的，而且很自然地，它们具有读取语法。

? (aref #0A5)
5