回复：常见问题解答20-12 [英] Re: FAQ 20-12
问题描述
7月3日，6：51 * pm，viza< tom.v ... @ gm-il.com.obviouschange.invalid>
写道：
On Jul 3, 6:51*pm, viza <tom.v...@gm-il.com.obviouschange.invalid>
wrote:
* * do {
* * * * * ** - p =" 0123456789abcdef" [num％base]; * LT;< -------这个
* *do {
* * * * * **--p = "0123456789abcdef"[num % base]; *<< -------This
line让我感到困惑？如果有人可以提供帮助，我会很感激。
line has me confused? If someone could help, I will appreciate it.
让我把它分开：
do {
* char hexidecimal_digits [ ] = {''0''，''1''，''2''，......等};
* p = p - 1;
* * p = hexidecimal_digits [num％base];
你可能不理解的唯一一点数学是（num％
base）表示num除以base时的余数。
Let me space it out a bit:
do {
* char hexidecimal_digits[]= {''0'',''1'',''2'', ...etc };
* p= p - 1;
* *p= hexidecimal_digits[ num % base ];
The only bit of maths that you might not understand is that ( num %
base ) means the remainder when num is divided by base.
我从未见过（这不一定意味着什么）一个
字符数组初始化。我很好奇你怎么从这个（0123456789abcdef[num％base];）到这个（char
hexidecimal_digits [] = {''0 ''，''1'，''2''，......等等;）。 （数学很好
（模数运算符）。谢谢你的帮助。
I have just never seen ( not that that necessarily means anything) a
char array initialized like that. I am curious as to how you get from
this ("0123456789abcdef"[num % base];) to this (char
hexidecimal_digits[]= {''0'',''1'',''2'', ...etc };). ( The math is fine
( viz the modulo operator). Thank you for your help.
推荐答案
mdh说：
mdh said:
7月3日下午6:51，viza< tom.v ... @ gm-il.com.obviouschange.invalid>
写道：
On Jul 3, 6:51 pm, viza <tom.v...@gm-il.com.obviouschange.invalid>
wrote:
do {
* - p =" 0123456789abcdef" [num％base];<< -------这个
do {
*--p = "0123456789abcdef"[num % base]; << -------This
>行让我感到困惑？如果有人可以帮助，我将不胜感激。
>line has me confused? If someone could help, I will appreciate it.
让我稍微说清楚：
做{\\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ r \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ n \\ necc ''，''1'，''2'，......等};
p = p - 1;
* p = hexidecimal_digits [num％你可能不理解的唯一一些数学是（num％
base）表示num除以base时的余数。
Let me space it out a bit:
do {
char hexidecimal_digits[]= {''0'',''1'',''2'', ...etc };
p= p - 1;
*p= hexidecimal_digits[ num % base ];
The only bit of maths that you might not understand is that ( num %
base ) means the remainder when num is divided by base.
我从来没有见过（这不一定意味着什么）一个初始化的
char数组。我很好奇你怎么从这个（0123456789abcdef[num％base];）到这个（char
hexidecimal_digits [] = {''0 ''，''1'，''2''，......等等;）。
I have just never seen ( not that that necessarily means anything) a
char array initialized like that. I am curious as to how you get from
this ("0123456789abcdef"[num % base];) to this (char
hexidecimal_digits[]= {''0'',''1'',''2'', ...etc };).
来吧迈克尔，他只是将它分散一点，以帮助你看到
逻辑，这就是全部。没有hexadecimal_digits数组，因为它不是必需的
- 字符串文字也可以。但是什么viza
告诉你的是我们可以编写具有相同效果的代码，使用
临时数组而不是字符串文字，你似乎同意
这个等价物很容易理解。
诀窍在于没有技巧。 A[0]是A。 AB[0]也是A，
和AB[1]是B。 ABC[0]是A，ABC[1]是B，ABC[2]是C。
" 0123456789abcdef" [0]为''0''。
" 0123456789abcdef" [1]为''1'。
" 0123456789abcdef" [9] is''9''。
" 0123456789abcdef" [10] is''a''。
" 0123456789abcdef" [11] is ''b''。
现在清除？
-
Richard Heathfield< http：/ /www.cpax.org.uk>
电子邮件：-http：// www。 + rjh @
谷歌用户：< http://www.cpax.org.uk/prg/writings/googly.php>
Usenet是一个奇怪的放置" - dmr 1999年7月29日
Come on Michael, he''s just spreading it out a little to help you see the
logic, that''s all. There is no hexadecimal_digits array, because it isn''t
necessary - the string literal will do just as well. But what viza is
telling you is that we could write code that has the same effect, using a
temporary array instead of a string literal, and you seem to agree that
this equivalent would be easy to understand.
The trick is that there is no trick. "A"[0] is ''A''. "AB"[0] is also ''A'',
and "AB"[1] is ''B''. "ABC"[0] is ''A'', "ABC"[1] is ''B'', "ABC"[2] is ''C''.
"0123456789abcdef"[0] is ''0''.
"0123456789abcdef"[1] is ''1''.
"0123456789abcdef"[9] is ''9''.
"0123456789abcdef"[10] is ''a''.
"0123456789abcdef"[11] is ''b''.
Clear now?
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
mdh写道：
mdh wrote:
>
>
.... snip ...
.... snip ...
>
我从来没有见过（这不一定意味着什么）
a char数组初始化。我很好奇你怎么从这个（0123456789abcdef[num％base];）到这个（char
hexidecimal_digits [] = {''0 ''，''1'，''2''，......等等;）。 （数学很好
（模数运算符）。谢谢你的帮助。
>
I have just never seen ( not that that necessarily means anything)
a char array initialized like that. I am curious as to how you get
from this ("0123456789abcdef"[num % base];) to this (char
hexidecimal_digits[]= {''0'',''1'',''2'', ...etc };). ( The math is fine
(viz the modulo operator). Thank you for your help.
为什么难度？012。 ..def"是一个char数组.num％base是一个在0到15范围内的
数字，只要num和base是
为正数，而base是范围1到16.因此
" 0123 ... def" [num％base]选择其中一个字符。
-
[mail]：Chuck F（cinefalconer at maineline dot net）
[page]：< http://cbfalconer.home.att.net>
尝试下载部分。
Why the difficulty? "012...def" is a char array. num % base is a
number in the range 0 through 15 as long as num and base are
positive, and base is in the range 1 through 16. Therefore
"0123...def"[num % base] selects one of those characters.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
7月3日，10：34 * pm，Richard Heathfield< r ... @ see.sig。 invalidwrote：
On Jul 3, 10:34*pm, Richard Heathfield <r...@see.sig.invalidwrote:
mdh说：
mdh said:
来吧Michael，他只是把它分散一点，以帮助你看到
逻辑，这就是全部。
Come on Michael, he''s just spreading it out a little to help you see the
logic, that''s all.
对我来说听起来很合乎逻辑！ />
没有hexadecimal_digits数组，因为它不是
Sounds perfectly logical to me!
There is no hexadecimal_digits array, because it isn''t
必需 - 字符串文字也可以。
necessary - the string literal will do just as well.
.... snip ....
....snip....
诀窍在于没有技巧。 A[0]是A。 AB[0]也是A，
和AB[1]是B。 ABC[0]是A，ABC[1]是B，ABC[2]是C。
" 0123456789abcdef" [0]为''0''。
" 0123456789abcdef" [1]为''1'。
" 0123456789abcdef" [9] is''9''。
" 0123456789abcdef" [10] is''a''。
" 0123456789abcdef" [11] is ''b''。
现在清除？
The trick is that there is no trick. "A"[0] is ''A''. "AB"[0] is also ''A'',
and "AB"[1] is ''B''. "ABC"[0] is ''A'', "ABC"[1] is ''B'', "ABC"[2] is ''C''.
"0123456789abcdef"[0] is ''0''.
"0123456789abcdef"[1] is ''1''.
"0123456789abcdef"[9] is ''9''.
"0123456789abcdef"[10] is ''a''.
"0123456789abcdef"[11] is ''b''.
Clear now?
是的......我已经回去看了K& R并同意这是一个非常巧妙的方式。我想说的是，即使
构造我是一个字符串文字在K& R中得到了很好的解释，我没有意识到人们可以通过使用
紧跟其后的运算符[]来引用常量中的每个字符。我所知道的是，当我在过去只是假装的时候，它经常让我进入
热水...所以我知道我可以在这里询问我所有的假设都将是
迅速纠正!!! :-)
谢谢理查德。
Yep...I have gone back and looked at K&R and agree it is a very neat
way of doing this. What I was trying to say was that even though the
construct "I am a string literal" is well explained in K&R, I was not
aware that one could refer to each character in the constant by using
the operator [] immediately following it. All I know is that when I
have simply assumed things in the past, it has often gotten me into
hot water...so I know I can ask here and all my assumptions will be
rapidly corrected!!! :-)
Thank you Richard.
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