为什么strcpy，strcat，& co返回一个指针？ [英] why do strcpy, strcat, & co return a pointer ?
问题描述
在标准C库的大多数实现中，
在先前分配的缓冲区中复制字符的函数，如strcpy或strcat，
返回指向该值的指针在我的NetBSD系统上，man strcpy给出了以下原型：
char * strcpy（char * restrict dst，const char * restrict src）;
在电话会议之前返回我们已经知道的指针有什么意义？
提前感谢您的解释。
On most implementations of the standard C library, the functions that
copy characters in a previously allocated buffer, like strcpy or strcat,
are returning the pointer to that buffer.
On my NetBSD system, man strcpy gives the following prototype :
char *strcpy(char * restrict dst, const char * restrict src);
What''s the point in returning a pointer we already know before the call?
Thank you in advance for an explanation.
推荐答案
" Nicolas" < ad*@serpe.org>在留言中写道
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"Nicolas" <ad*@serpe.org> wrote in message
news:3f***********************@news.free.fr...
在标准C库的大多数实现中，在先前分配的缓冲区中复制字符的函数，如strcpy或strcat，
返回指向该缓冲区的指针。
<在我的NetBSD系统上，man strcpy给出了以下原型：
char * strcpy（char * restrict dst，const char * restrict src）;
有什么意义在电话会议之前返回我们已经知道的指针？
提前感谢您的解释。
On most implementations of the standard C library, the functions that
copy characters in a previously allocated buffer, like strcpy or strcat,
are returning the pointer to that buffer.
On my NetBSD system, man strcpy gives the following prototype :
char *strcpy(char * restrict dst, const char * restrict src);
What''s the point in returning a pointer we already know before the call?
Thank you in advance for an explanation.
这样你就可以写（不必要的缩写） ）这样的代码：
strcat（path，strcpy（file，" fred.txt"））;
So that you can write (unnecessarily abbreviated) code like this:
strcat(path, strcpy(file, "fred.txt"));
" Nicolas" < ad*@serpe.org>在留言中写道
新闻：3f *********************** @ news.free.fr ...
"Nicolas" <ad*@serpe.org> wrote in message
news:3f***********************@news.free.fr...
在标准C库的大多数实现中，在先前分配的缓冲区中复制字符的函数，如strcpy或strcat，
返回指向该缓冲区的指针。
<在我的NetBSD系统上，man strcpy给出了以下原型：
char * strcpy（char * restrict dst，const char * restrict src）;
有什么意义在通话之前返回我们已经知道的指针？
On most implementations of the standard C library, the functions that
copy characters in a previously allocated buffer, like strcpy or strcat,
are returning the pointer to that buffer.
On my NetBSD system, man strcpy gives the following prototype :
char *strcpy(char * restrict dst, const char * restrict src);
What''s the point in returning a pointer we already know before the call?
方便。其中一个函数调用的结果可能会被用作更大表达式的一部分，例如：
printf（"％s \ n" ，strcat（s1，s2））;
如果您不需要返回值，请忽略它。
strcat（ s1，s2）;
-Mike
Convenience. The result of one of these function calls might
be used as part of a larger expression, e.g.:
printf("%s\n", strcat(s1, s2));
If you don''t need the return value, just ignore it.
strcat(s1, s2);
-Mike
Nicolas写道：
Nicolas wrote:
关于大多数标准C库的实现，在先前分配的缓冲区中复制字符的函数，比如strcpy
或strcat，都会返回指向该缓冲区的指针。
在我的NetBSD系统，man strcpy给出了以下原型：
char * strcpy（char * restrict dst，const char * restrict src）;
返回指针的重点是什么
电话之前知道吗？
On most implementations of the standard C library, the functions
that copy characters in a previously allocated buffer, like strcpy
or strcat, are returning the pointer to that buffer.
On my NetBSD system, man strcpy gives the following prototype :
char *strcpy(char * restrict dst, const char * restrict src);
What''s the point in returning a pointer we already know before the
call?
它让您对以下特有的运行时动作感到困惑：
char p [100] =" Fred";
....
printf（"％s或％s \ n"，p，strcat（p，"和George＃））;
对于改进的语义，查找（非标准）strlcat和
strlcpy函数。
-
Chuck F（cb********@yahoo.com）（cb ******** @ worldnet.att.net）
可用于咨询/临时嵌入式和系统。
< http://cbfalconer.home.att.net>使用worldnet地址！
It allows you to be baffled by the peculiar run-time actions of:
char p[100] = "Fred";
....
printf("%s or %s\n", p, strcat(p, " and George"));
For improved semantics look up the (non-standard) strlcat and
strlcpy functions.
--
Chuck F (cb********@yahoo.com) (cb********@worldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net> USE worldnet address!
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