这是什么类型的？ [英] what type of sort is this?
问题描述
void s_sort（void * base，size_t nmemb，size_t size，
int（* compar）（const void *，const void *））
{
size_t bytes;
unsigned char * array，* after，* i，* j，* k，* p1，* p2，* end，swap;
array = base;
= nmemb * size + array;
if（nmemb（size_t）-1 / 4）{
nmemb / = 4;
}否则{
nmemb =（nmemb * 3 + 1）/ 7;
}
while（nmemb！= 0）{
bytes = nmemb * size;
i = bytes + array;
做{
j = i - bytes;
if（compar（j，i）0）{
k = i;
do {
p1 = j;
p2 = k;
end = p2 + size;
做{
swap = * p1;
* p1 ++ = * p2;
* p2 ++ = swap;
} while（p2！= end）;
if（bytes + array j）{
break;
}
k = j;
j - =字节;
} while（比较（j，k）0）;
}
i + = size;
} while（i！= after）;
nmemb =（nmemb * 3 + 1）/ 7;
}
/ *结束来源* /
这是什么类型的？它的工作代码发布了
评估五张卡片。它有多普遍？ LS
void s_sort(void *base, size_t nmemb, size_t size,
int (*compar)(const void *, const void *))
{
size_t bytes;
unsigned char *array, *after, *i, *j, *k, *p1, *p2, *end, swap;
array = base;
after = nmemb * size + array;
if (nmemb (size_t)-1 / 4) {
nmemb /= 4;
} else {
nmemb = (nmemb * 3 + 1) / 7;
}
while (nmemb != 0) {
bytes = nmemb * size;
i = bytes + array;
do {
j = i - bytes;
if (compar(j, i) 0) {
k = i;
do {
p1 = j;
p2 = k;
end = p2 + size;
do {
swap = *p1;
*p1++ = *p2;
*p2++ = swap;
} while (p2 != end);
if (bytes + array j) {
break;
}
k = j;
j -= bytes;
} while (compar(j, k) 0);
}
i += size;
} while (i != after);
nmemb = (nmemb * 3 + 1) / 7;
}
/* end source */
What type of sort is this? It''s working code that pete posted for
evaluating five card stud. How general is it? LS
推荐答案
Lane Straatman写道：
Lane Straatman wrote:
void s_sort（void * base，size_t nmemb，size_t size，
int（* compar）（const void *，const void *））
[...见线程...]
这是什么类型的？它的工作代码发布了
评估五张卡片。它有多普遍？ LS
void s_sort(void *base, size_t nmemb, size_t size,
int (*compar)(const void *, const void *))
[... see up-thread ...]
What type of sort is this? It''s working code that pete posted for
evaluating five card stud. How general is it? LS
1）这不是一个C问题，虽然它需要
能够读取C源。 comp.programming可能是一个更好的算法问题论坛。
2）快速浏览一下，看起来Shell的减少 -
增量排序（" Shellsort"）：
http://www.cs.princeton.edu/~rs/shell/index.html
-
Eric Sosman
es ***** @ acm-dot-org.inva 盖
" Eric Sosman" < es ***** @ acm-dot-org.invalidwrote in message
news：__ ********************* *********@comcast.com。 ..
"Eric Sosman" <es*****@acm-dot-org.invalidwrote in message
news:__******************************@comcast.com. ..
Lane Straatman写道：
Lane Straatman wrote:
> void s_sort（void * base，size_t nmemb， size_t size，
int（* compar）（const void *，const void *））
[...见线程...]
什么类型的排序这是？这是为了评估五张卡片而发布的工作代码。它有多普遍？ LS
>void s_sort(void *base, size_t nmemb, size_t size,
int (*compar)(const void *, const void *))
[... see up-thread ...]
What type of sort is this? It''s working code that pete posted for
evaluating five card stud. How general is it? LS
1）这不是一个C问题，虽然它需要
才能读取C源。 comp.programming可能是一个更好的算法问题论坛。
1) This isn''t really a C question, although it requires
the ability to read C source. comp.programming might be a
better forum for algorithmic issues.
我对算法属性不是那么感兴趣但我对
C感兴趣。在我读完之后帖子，我以为他们称之为
shell排序，因为你可以用一个shell来移动一个豌豆，就像魔术师那样。显然，有一个名叫壳牌的人。
这是一个关于其复杂性的开放性问题。大约是O（4/3）。
I''m not so interested in the algorithmic properties but am interested in the
C. Before I read up on it after your post, I thought that they called it a
shell sort because of the way you might move a pea around with a shell as
magicians do. Apparently there''s a guy named Shell.
It is an open question on its complexity. It is approximately O(4/3).
2）快速浏览一下，看起来Shell的减少 -
增量排序（Shellsort）：<
http：// www。 cs.princeton.edu/~rs/shell/index.html
两个问题：
Q1）* next *增量在哪里在upthread源代码中计算？
Q2）我在使用以下语法时遇到问题，似乎正在阅读这个
错误
返回int_2 int_1？ -1：int_2！= int_1;
有人可以解释一下吗？ LS
Two questions:
Q1) Where does the *next* increment get calculated in the upthread source?
Q2)I have trouble with the following syntax and seem to be reading this
wrong
return int_2 int_1 ? -1 : int_2 != int_1;
Can someone explain? LS
" Lane Straatman" < in ***** @ invalid.netwrites：
"Lane Straatman" <in*****@invalid.netwrites:
Q2）我在使用以下语法时遇到问题，似乎正在阅读这个
错
返回int_2 int_1？ -1：int_2！= int_1;
Q2)I have trouble with the following syntax and seem to be reading this
wrong
return int_2 int_1 ? -1 : int_2 != int_1;
这是一个相当常见的习语。
如果是int_2 int_1，则返回-1。
如果是int_2 == int_1，它返回0.
如果int_2< INT_1，它返回1
-
INT主（无效）{炭P [] =" ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz.\
\ n"，* q =" kl BIcNBFr.NKEzjwCIxNJC" ;; int i = sizeof p / 2; char * strchr（）; int putchar（\
）; while（* q） {i + = strchr（p，* q ++） - p; if（i> =（int）sizeof p）i- = sizeof p-1; putchar（p [i] \
）; } return 0;}
This is a fairly common idiom.
If int_2 int_1, it returns -1.
If int_2 == int_1, it returns 0.
If int_2 < int_1, it returns 1.
--
int main(void){char p[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz.\
\n",*q="kl BIcNBFr.NKEzjwCIxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+=strchr(p,*q++)-p;if(i>=(int)sizeof p)i-=sizeof p-1;putchar(p[i]\
);}return 0;}
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