strtok上的段错误 [英] segfault on strtok
问题描述
..以下会导致段错误。 ..没有
知道为什么？！
int main（）{
char name [15];
strcpy（name，" ab8bc8cd8ed"）;
char cur [800];
strcpy（cur，strtok（name） ，8））;
while（cur）{
printf（" Output：％s \ n"，cur）;
printf（" Stringlength％i \ n"，strlen（cur））;
strcpy（cur，strtok（0，" 8"））;
printf（" next \ n"）;
}
}
输出：-----------------------
输出：ab
Stringlength 2
next
输出：bc
Stringlength 2
next
输出： cd
Stringlength 2
next
输出：ed
Stringlength 2
Segmentationfault
------------------------------
感谢您的帮助
问候
Fatih Gey
在文章< bn ************* @ news.t-online.com>中，Fatih Gey写道：
[cut ]
#include< stdio.h>
#include< string.h>
int main（）{name char [15];
strcpy（name，" ab8bc8cd8ed"）;
char cur [800];
strcpy（cur， strtok（name，" 8"））;
while（cur）{
cur永远不会为零。
printf（"输出：％s \ n"，cur）;
printf（" Stringlength％i \ n"，strlen（cur））;
strcpy（cur，strtok（ 0，8））;
（严格来说，该行的零点应该是空的
指针）
printf（" next \ n"）;
}
}
检查strtok（）的返回值。当没有
更多令牌时，它返回NULL。 NULL在< stdlib.h>中定义。
干杯，
Andreas
-
Andreas K？h？ri
Andreas Kahari写道：文章< bn ********** ***@news.t-online.com>，Fatih Gey写道：
printf（" Output：％s \ n"，cur）;
printf （Stringlength％i \ n，strlen（cur））;
strcpy（cur，strtok（0，" 8"））;
（严格来说）说来，该行上的零应该是一个空的
指针）
0转换为指针*是*空指针。
在这里使用0没有错。
-Kevin
-
我的电子邮件地址有效，但会定期更改。
要联系我，请使用最近发布的地址。
Andreas Kahari写道：
在文章< bn ************* @ news.t-online.com>中，Fatih Gey写道：
[cut]
#include< stdio.h>
#include< string.h>
NULL在< stdlib.h>中定义。
NULL也在stdio.h和string.h中定义，
以及其他头文件。
stddef.h有NULL宏和_t typedef。
NULL和所有其他宏和typedef，
也在每个头文件中定义，对于任何函数都有原型
，定义为
能够接受他们[NULL和朋友]，作为参数，
或将它们作为值返回。
-
pete
Hi,
.. following causes a segfault. .. didn''t
know why ?!
int main() {
char name[15];
strcpy (name, "ab8bc8cd8ed");
char cur[800];
strcpy (cur, strtok(name, "8"));
while (cur) {
printf ("Output: %s\n", cur);
printf ("Stringlength %i\n", strlen(cur));
strcpy(cur,strtok(0, "8"));
printf ("next\n");
}
}
output: -----------------------
Output: ab
Stringlength 2
next
Output: bc
Stringlength 2
next
Output: cd
Stringlength 2
next
Output: ed
Stringlength 2
Segmentationfault
------------------------------
Thanks for any help
Regards
Fatih Gey
In article <bn*************@news.t-online.com>, Fatih Gey wrote:
[cut]
#include <stdio.h>
#include <string.h>
int main() {
char name[15];
strcpy (name, "ab8bc8cd8ed");
char cur[800];
strcpy (cur, strtok(name, "8"));
while (cur) {
cur is never going to be zero.
printf ("Output: %s\n", cur);
printf ("Stringlength %i\n", strlen(cur));
strcpy(cur,strtok(0, "8"));
(strictly speaking, the zero on that line should be a null
pointer)
printf ("next\n");
}
}
Check the return value of strtok(). It returns NULL when no
more tokens remains. NULL is defined in <stdlib.h>.
Cheers,
Andreas
--
Andreas K?h?ri
Andreas Kahari wrote:In article <bn*************@news.t-online.com>, Fatih Gey wrote:printf ("Output: %s\n", cur);
printf ("Stringlength %i\n", strlen(cur));
strcpy(cur,strtok(0, "8"));
(strictly speaking, the zero on that line should be a null
pointer)
0 converted to a pointer *is* a null pointer. There''s nothing wrong with
using 0 here.
-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
Andreas Kahari wrote:
In article <bn*************@news.t-online.com>, Fatih Gey wrote:
[cut]
#include <stdio.h>
#include <string.h> NULL is defined in <stdlib.h>.
NULL is also defined in stdio.h and string.h,
as well as in other header files.
stddef.h has the NULL macro and the _t typedefs.
NULL and all of the other macros and typedefs,
are also defined in each header file that has a prototype
for any function, which is defined as either
being able to accept them [NULL and friends], as arguments,
or return them as values.
--
pete
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