双重Jeopoardy [英] Double Jeopoardy

问题描述

d1和d2是我们可以存储为双打（8字节）的数字。


d1> 0

d2> 0

d1< d2


现在s1和s2是长度为8个字节的字符串。 （我们可以在这里忽略unicode）。


进入s1和s2我们分别从d1和d2复制8个字节。


是否遵循s1< s2，（或s1在s2之前订购）在按

文本或二进制比较排序时？


为什么？


为什么我问？好吧，如果我们可以证明这是如此（s1< s2 => d1< d2）那么

我们可以通过使用Wizhook.SortStringArray订购数组数组

方法。


我为什么要这样做？ ......没有特别的原因......我有一堆

其他分拣程序。


-

Lyle

（电子邮件参考 http://ffdba.com/ contacts.htm ）

d1 and d2 are numbers we can store as doubles (8 bytes).

d1 > 0
d2 > 0
d1 < d2

Now s1 and s2 are strings of length 8 bytes. (We can ignore unicode here).

Into s1 and s2 we copy the 8 bytes from d1 and d2 respectively.

Does it follow that s1<s2, (or s1 is ordered before s2) when sorted on a
text or binary compare?

Why?

Why do I ask? Well, if we could prove that this is so (s1<s2 => d1<d2) then
we could order arrays of numbers by using the Wizhook.SortStringArray
method.

Why would I want to do that? ... No particular reason ... I have a bunch of
other sorting procedures.

--
Lyle
(for e-mail refer to http://ffdba.com/contacts.htm)

推荐答案

WizHook在A03中，所以即使它已经使用了一段时间也是安全的隐藏

和无证件。


有趣的是在这里谈论它：
http://www.utteraccess.com/forums/ac...ess561560.html


-

Jerry Boone

Analytical Technologies，Inc。
http://www.antech.biz

ASP，ASP.NET，SQL Server和
的安全托管和开发解决方案
访问


" Lyle Fairfield" <弥************ @ Invalid.Com>在消息中写道

news：Xn ****************** @ 130.133.1.4 ...

WizHook is in A03 so it''s safe to use for a while even though it''s hidden
and undocumented.

Interesting talk about it here though:
http://www.utteraccess.com/forums/ac...ess561560.html

--
Jerry Boone
Analytical Technologies, Inc.
http://www.antech.biz
Secure Hosting and Development Solutions for ASP, ASP.NET, SQL Server, and
Access

"Lyle Fairfield" <Mi************@Invalid.Com> wrote in message
news:Xn******************@130.133.1.4...

d1和d2是我们可以存储为双打（8字节）的数字。

d1> 0
d2> 0
d1< d2
现在s1和s2是长度为8个字节的字符串。 （我们可以在这里忽略unicode）。

进入s1和s2我们分别从d1和d2复制8个字节。

是否遵循s1< s2，（或s1）是在s2之前订购的，当按照文本或二进制比较排序时？

为什么？

我为什么要问？好吧，如果我们可以证明这是如此（s1< s2 => d1< d2）
那么我们可以通过使用Wizhook.SortStringArray
方法来排序数字数组。
的其他分拣程序。

-
Lyle
（电子邮件参考 http://ffdba.com/contacts.htm ）

d1 and d2 are numbers we can store as doubles (8 bytes).

d1 > 0
d2 > 0
d1 < d2

Now s1 and s2 are strings of length 8 bytes. (We can ignore unicode here).

Into s1 and s2 we copy the 8 bytes from d1 and d2 respectively.

Does it follow that s1<s2, (or s1 is ordered before s2) when sorted on a
text or binary compare?

Why?

Why do I ask? Well, if we could prove that this is so (s1<s2 => d1<d2) then we could order arrays of numbers by using the Wizhook.SortStringArray
method.

Why would I want to do that? ... No particular reason ... I have a bunch of other sorting procedures.

--
Lyle
(for e-mail refer to http://ffdba.com/contacts.htm)

Lyle Fairfield写道：

Lyle Fairfield wrote:

d1和d2是我们可以存储为双打（8字节）的数字。

d1> 0
d2> 0
d1< d2
现在s1和s2是长度为8个字节的字符串。 （我们可以在这里忽略unicode）。

进入s1和s2我们分别从d1和d2复制8个字节。

是否遵循s1< s2，（或s1）是在s2之前订购的，当按照文本或二进制比较排序时？

为什么？

我为什么要问？好吧，如果我们能够证明这是如此（s1< s2 => d1< d2）那么我们可以通过使用Wizhook.SortStringArray
方法来排序数字数组。
其他分拣程序。

- Lyle
（电子邮件参考 http://ffdba.com/contacts.htm ）

d1 and d2 are numbers we can store as doubles (8 bytes).

d1 > 0
d2 > 0
d1 < d2

Now s1 and s2 are strings of length 8 bytes. (We can ignore unicode here).

Into s1 and s2 we copy the 8 bytes from d1 and d2 respectively.

Does it follow that s1<s2, (or s1 is ordered before s2) when sorted on a
text or binary compare?

Why?

Why do I ask? Well, if we could prove that this is so (s1<s2 => d1<d2) then
we could order arrays of numbers by using the Wizhook.SortStringArray
method.

Why would I want to do that? ... No particular reason ... I have a bunch of
other sorting procedures.

--
Lyle
(for e-mail refer to http://ffdba.com/contacts.htm)

因为它们是不同的存储介质（例如......你获得两倍的前3个字节的多少次？）我建议你将它们存储到一个字符串左边

填充。由于该方法增加了开销，我怀疑我会使用它。


我们已经看到1,10,100在2,20,200之前的字符串排序...... .unless

填充。

Since they are different storage mediums (ex...how many times do you get the
first 3 bytes of a double?) I would suggest you store them into a string left
padded. Since that method adds overhead, I doubt I''d use it.

We''ve seen where 1,10,100 come before 2, 20, 200 in a string sort....unless
padded.

Salad< oi*@vinegar.com>在新闻中写道：40 *************** @ vinegar.com:

Salad <oi*@vinegar.com> wrote in news:40***************@vinegar.com:

Lyle Fairfield写道：

Lyle Fairfield wrote:

d1和d2是我们可以存储为双打（8字节）的数字。

d1> 0
d2> 0
d1< d2
现在s1和s2是长度为8个字节的字符串。 （我们可以忽略unicode
这里）。

进入s1和s2我们分别从d1和d2复制8个字节。

是否遵循s1< s2 ，（或s1在s2之前订购）当按文本或二进制比较排序时？

为什么？

我为什么要问？好吧，如果我们能够证明这是如此（s1< s2 => d1< d2）
那么我们可以使用
Wizhook.SortStringArray方法对数字数组进行排序。

-
Lyle
（电子邮件参考 http://ffdba.com/contacts.htm ）

d1 and d2 are numbers we can store as doubles (8 bytes).

d1 > 0
d2 > 0
d1 < d2

Now s1 and s2 are strings of length 8 bytes. (We can ignore unicode
here).

Into s1 and s2 we copy the 8 bytes from d1 and d2 respectively.

Does it follow that s1<s2, (or s1 is ordered before s2) when sorted on
a text or binary compare?

Why?

Why do I ask? Well, if we could prove that this is so (s1<s2 => d1<d2)
then we could order arrays of numbers by using the
Wizhook.SortStringArray method.

Why would I want to do that? ... No particular reason ... I have a
bunch of other sorting procedures.

--
Lyle
(for e-mail refer to http://ffdba.com/contacts.htm)

因为它们是不同的存储介质（例如...你得到多少次双倍的前三个字节？）我建议你将它们存储在一个左边填充的字符串中。由于该方法增加了开销，我怀疑我会使用
它。

我们已经看到1,10,100来自2,20,200之前的字符串
排序....除非填充。

Since they are different storage mediums (ex...how many times do you get
the first 3 bytes of a double?) I would suggest you store them into a
string left padded. Since that method adds overhead, I doubt I''d use
it.

We''ve seen where 1,10,100 come before 2, 20, 200 in a string
sort....unless padded.

叹息......


-

Lyle

（电子邮件参考 http：// ffdba。 com / contacts.htm ）

这篇关于双重Jeopoardy的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！