这个定义明确吗？ [英] Is this well-defined?
问题描述
您好。我一直在使用这样的代码：
* p ++ =''x'';
* p ++ = * q ++;
现在很长一段时间了。有用。下面的代码编译没有
警告并正确运行。夹板也不会抱怨
这些线。
这是根据标准定义的吗？关于序列点，我读了一点
，而没有像i = i ++那样做。这个
似乎是相似的。如果您正在使用和
修改* same *变量，不止一次，没有
干预序列点，这只是一个问题吗？
特别是，
* p ++ = * q ++;
保证取消引用p和q，并执行作业，
之前是否增加了？
---------------------- ----------------------------
#include< stdio.h>
int main（无效）
{
char s [] =" some\xt\\\
in\\\
several\\\
lines\ n" ;;
char t [100]，* src，* dest;
//用CRLF替换每个LF
for（src = s，dest = t; * src;）{
if（* src ==''\ n''）
* dest ++ ='' \'';;
* dest ++ = * src ++;
}
* dest =''\ 0'';
fputs（t，stdout）;
返回0;
}
----- ------------------------------ ---------------
谢谢，
-
Tom Zych
这个电子邮件地址将在某个时候到期以阻止垃圾邮件发送者。
永久地址：echo''g******@cbobk.pbz''| rot13
Hi. I''ve been using code like this:
*p++ = ''x'';
*p++ = *q++;
for a long time now. It works. The code below compiles with no
warnings and runs correctly. Even splint doesn''t complain about
those lines.
Is this well-defined according to the standard? I''ve read a little
about sequence points and not doing things like "i = i++". This
would seem to be similar. Is it only a problem if you''re using and
modifying the *same* variable, more than once, without an
intervening sequence point?
In particular, is
*p++ = *q++;
guaranteed to dereference p and q, and perform the assignment,
before either is incremented?
--------------------------------------------------
#include <stdio.h>
int main(void)
{
char s[] = "Some\ntext\nin\nseveral\nlines\n";
char t[100], *src, *dest;
// replace each LF with CRLF
for (src = s, dest = t; *src; ) {
if (*src == ''\n'')
*dest++ = ''\r'';
*dest++ = *src++;
}
*dest = ''\0'';
fputs(t, stdout);
return 0;
}
--------------------------------------------------
Thanks,
--
Tom Zych
This email address will expire at some point to thwart spammers.
Permanent address: echo ''g******@cbobk.pbz'' | rot13
推荐答案
Tom Zych< tz ****** @ pobox.com>在新闻中写道：3F *************** @ pobox.com：
Tom Zych <tz******@pobox.com> wrote in news:3F***************@pobox.com:
嗨。我一直在使用这样的代码：
* p ++ =''x'';
* p ++ = * q ++;
很好。
现在很久了。有用。下面的代码编译没有
警告并正确运行。夹板也不会抱怨这些线条。
当然。
这是根据标准定义的吗？我读了一些关于序列点的内容，而没有做过像i = i ++这样的事情。这似乎是相似的。如果您在不使用
干预序列点的情况下使用和修改* same *变量，这只是一个问题吗？
特别是<
* p ++ = * q ++;
保证取消引用p和q，然后执行赋值，
再增加之前？
Hi. I''ve been using code like this:
*p++ = ''x'';
*p++ = *q++;
Fine.
for a long time now. It works. The code below compiles with no
warnings and runs correctly. Even splint doesn''t complain about
those lines.
Sure.
Is this well-defined according to the standard? I''ve read a little
about sequence points and not doing things like "i = i++". This
would seem to be similar. Is it only a problem if you''re using and
modifying the *same* variable, more than once, without an
intervening sequence point?
In particular, is
*p++ = *q++;
guaranteed to dereference p and q, and perform the assignment,
before either is incremented?
是的，这是明确定义的。必须计算* p和* q的值，并使用
作为分配。此外，p和q必须递增。这个
发生的顺序取决于编译器，但结果是明确定义的，是你期望的
。
-
- 马克 - >
-
Yes this is well defined. The values of *p and *q must be calculated and
used for the assigment. Also, p and q must be incremented. The order this
happens is up to the compiler but the result is well-defined and is what
you expect.
--
- Mark ->
--
Tom Zych写道：
Tom Zych wrote:
特别是，
* p ++ = * q ++;
保证取消引用p和q，然后执行赋值，
增加了吗？
In particular, is
*p++ = *q++;
guaranteed to dereference p and q, and perform the assignment,
before either is incremented?
号但是谁在乎呢？与i = i ++不同，上面的语句没有两次修改_same_位置的
。也就是说，除非p或q指向他们自己
（投射到appropritate类型） - 这将是一件奇怪的事情。
您的代码是很好。
好吧，我会使用const char src []和const char * s，或者只是使用
char * s =" Some \
text\\\
in\\\
several\\\
lines\\\
英寸;并且删除了src，这样当字符串文字支持
时，字符串文字最终会进入只读内存。和/ ** / comments而不是//注释如果代码是在B89编译器上编译的
。但这些是不同的问题。
-
Hallvard
No. But who cares? Unlike i = i++, the above statement doesn''t modify
the _same_ location twice. That is, unless p or q point to themselves
(cast to the appropritate type) - which would be a weird thing to do.
Your code is fine.
Well, I would have used const char src[] and const char *s, or just used
char *s = "Some\ntext\nin\nseveral\nlines\n"; and dropped src, so that
the string literal would end up in read-only memory when that is
supported. And /**/ comments instead of // comments if the code is
to be compiled on C89 compilers. But these are different issues.
--
Hallvard
Tom Zych< tz * *****@pobox.com>写道：
Tom Zych <tz******@pobox.com> wrote:
嗨。我一直在使用这样的代码：
* p ++ =''x'';
* p ++ = * q ++;
很长一段时间了。有用。下面的代码编译没有
警告并正确运行。夹板也不会抱怨这些线条。
这是根据标准定义的吗？
是的。
我读了一些关于序列点的内容，而没有做过像i = i ++这样的事情。这似乎是相似的。如果您在不使用
干预序列点的情况下多次使用和修改* same *变量，这只是一个问题吗？
确切地说。
特别是，
* p ++ = * q ++;
保证取消引用p和q，并执行分配，
之前是否递增？
是的。
------------------------------------ --------------
#include< stdio.h>
int main（void）
{
char s [] =" Some\xt\\\
in\\\
several\\\
lines\\\
" ;;
char t [100]，* src，* dest;
//替换每个LF与CRLF
for（src = s，dest = t; * src;）{
if（* src ==''\ n''）
* dest ++ ='' \'';
* dest ++ = * src ++;
}
* dest =''\ 0'';
fputs（t，stdout） ;
返回0;
}
--------------------------------- -----------------
看起来不错。
谢谢，
Hi. I''ve been using code like this:
*p++ = ''x'';
*p++ = *q++;
for a long time now. It works. The code below compiles with no
warnings and runs correctly. Even splint doesn''t complain about
those lines.
Is this well-defined according to the standard? Yes.
I''ve read a little
about sequence points and not doing things like "i = i++". This
would seem to be similar. Is it only a problem if you''re using and
modifying the *same* variable, more than once, without an
intervening sequence point? Exactly.
In particular, is
*p++ = *q++;
guaranteed to dereference p and q, and perform the assignment,
before either is incremented? Yes.
--------------------------------------------------
#include <stdio.h>
int main(void)
{
char s[] = "Some\ntext\nin\nseveral\nlines\n";
char t[100], *src, *dest;
// replace each LF with CRLF
for (src = s, dest = t; *src; ) {
if (*src == ''\n'')
*dest++ = ''\r'';
*dest++ = *src++;
}
*dest = ''\0'';
fputs(t, stdout);
return 0;
}
-------------------------------------------------- Looks good.
Thanks,
问候
Irrwahn
-
闭上眼睛，按三次逃生。
Regards
Irrwahn
--
Close your eyes and press escape three times.
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