strcpy问题 [英] strcpy question
问题描述
以下代码导致Segementation fault：
char * s1;
char * s2;
s2 =" HELLO";
strcpy（s1，s2）;
如果我将代码修改为：
char * s1;
char * s2;
s2 =" HELLO";
s1 =（char *）malloc（10 * sizeof（char））;
strcpy（s1，s2）;
代码效果很好。所以我应该总是为s1分配内存，然后
我称之为strcpy，对吧？
谢谢。
The following code causes "Segementation fault":
char *s1;
char *s2;
s2 = "HELLO";
strcpy(s1, s2);
If I modify the code to:
char *s1;
char *s2;
s2 = "HELLO";
s1 = (char*)malloc(10 * sizeof(char));
strcpy(s1, s2);
The code works well. So I should always allocate memory for s1, before
I call strcpy, right?
Thanks.
推荐答案
在文章< 11 ********************* @ y43g2000cwc.googlegroups中。 com>，
杰克< ju ****** @ gmail.com>写道：
In article <11*********************@y43g2000cwc.googlegroups. com>,
Jack <ju******@gmail.com> wrote:
以下代码导致Segementation fault：
char * s1;
char * s2;
s2 =" HELLO" ;;
strcpy（s1，s2）;
你已经声明s1是一个指针，但你没有给它
任何指向的内存。
如果我将代码修改为：
char * s1;
char * s2;
s2 =" HELLO";
s1 =（char *）malloc（10 * sizeof（char））;
strcpy（s1，s2）;
代码运行良好。所以我应该总是为s1分配内存，之后我会调用strcpy，对吗？
The following code causes "Segementation fault": char *s1;
char *s2; s2 = "HELLO"; strcpy(s1, s2);
You have declared s1 to be a pointer, but you haven''t given it
any memory to point to.
If I modify the code to:
char *s1;
char *s2;
s2 = "HELLO";
s1 = (char*)malloc(10 * sizeof(char));
strcpy(s1, s2);
The code works well. So I should always allocate memory for s1, before
I call strcpy, right?
不一定。例如，
char s1 [128];
char * s2 =" HELLO";
if（strlen（ s2）=> sizeof s1）fprintf（stderr，Opps，s2太长了\ n）
else strcpy（s1，s2）;
关键是目标存储必须存在才能写入
。 malloc（）是获取存储空间的一种方式，但是你也可以将
写入自动变量或静态变量，只要你确定你这样做就可以了。
不写入结束变量。
-
法律 - 它是一种商品
- Andrew Ryan（环球邮报，2005/11/26）
Not necessarily. For example,
char s1[128];
char *s2 = "HELLO";
if (strlen(s2) => sizeof s1) fprintf(stderr, "Opps, s2 is too long\n")
else strcpy(s1,s2);
The key is that the destination storage must exist before you can write
into it. malloc() is one way of getting that storage, but you can also
write into auto or static variables as long as you make sure you do
not write past the end of the variable.
--
"law -- it''s a commodity"
-- Andrew Ryan (The Globe and Mail, 2005/11/26)
杰克说：
Jack said:
以下代码导致Segementation fault：
char * s1;
char * s2;
s2 =" HELLO";
strcpy（s1，s2）;
是的，它会。
如果我将代码修改为：
char * s1;
char * s2;
s2 =" HELLO";
s1 =（char *）malloc（10 * sizeof（char））;
s1 = malloc（strlen（s2）+ 1）;
if（s1！= NULL）
{
strcpy（s1，s2）;
代码运行良好。所以我应该总是为s1分配内存，然后我打电话给strcpy，对吗？
The following code causes "Segementation fault":
char *s1;
char *s2;
s2 = "HELLO";
strcpy(s1, s2);
Yeah, it would.
If I modify the code to:
char *s1;
char *s2;
s2 = "HELLO";
s1 = (char*)malloc(10 * sizeof(char));
s1 = malloc(strlen(s2) + 1);
if(s1 != NULL)
{
strcpy(s1, s2);
The code works well. So I should always allocate memory for s1, before
I call strcpy, right?
是的。一切都必须在某个地方。你必须有一个地方可以放置它，
才能把它放在那里。
-
Richard Heathfield
Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk
电子邮件：rjh在上面的域名（但显然放弃了www）
Yeah. Everything Has To Be Somewhere. You have to have somewhere to put it,
before you can put it there.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
杰克写于06/01/06 16：28，：
Jack wrote On 06/01/06 16:28,:
以下代码导致Segementation fault：
char * s1;
char * s2;
s2 =" HELLO"
strcpy（s1，s2）;
如果我将代码修改为：
char * s1;
char * s2;
s2 =" HELLO";
s1 =（char *）malloc（ 10 * sizeof（char））;
strcpy（s1，s2）;
代码运行良好。所以我应该总是为s1分配内存，然后再调用strcpy，对吧？
The following code causes "Segementation fault":
char *s1;
char *s2;
s2 = "HELLO";
strcpy(s1, s2);
If I modify the code to:
char *s1;
char *s2;
s2 = "HELLO";
s1 = (char*)malloc(10 * sizeof(char));
strcpy(s1, s2);
The code works well. So I should always allocate memory for s1, before
I call strcpy, right?
请参阅comp.lang中的问题7.1,7.2和7.3b .c
常见问题（FAQ）列表
http://www.c-faq.com/
-
Er ********* @ sun.com
See Questions 7.1, 7.2, and 7.3b in the comp.lang.c
Frequently Asked Questions (FAQ) list
http://www.c-faq.com/
--
Er*********@sun.com
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