关于\0的问题 [英] Question on \0
问题描述
char * s =''\ 0''
if（* s ==''\'''）{
printf（" yes" ;）;
}
上述代码是否可以正常工作？
cr*******@gmail.com 说：
char * s =''\ 0''
if（* s ==''\'''）{
printf（是）;
}
上述代码是否可以正常运行？
不，它甚至不会编译。鉴于它只是一个代码片段，我们将会b / b忽略缺乏家具。但是第一个
线上缺少分号是不可能的，是吗？此外，虽然在这种特殊情况下，
值''\ 0''实际上/工作/作为指针的初始化（因为
它只是一个int值为0），会更合适。另外，
你会想要制作const，以防止意外写作
它。并且你需要确保你的文本输出流是正确的
用换行符终止：
#include< stdio.h>
int main（无效）
{
const char * s ="" ;;
if （* s ==''\'''）
{
printf（" yes \ n"）;
}
返回0;
}
-
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）
< blockquote class =post_quotes>
char * s =''\ 0''
if（* s ==''\'''）{
printf（是）;
}
好对不起，忘记了分号。我很抱歉，如果你看到
普通代码添加结尾''\ n''和main（）并且
$ b $不准确b什么不是 - 我在概念上要求更多，而不是实际使用
代码。我正在查看给我的
项目的框架代码，我相信我发现了一个bug，所以我想问一个
简化的问题，而不是给整个部分。
cr ******* @ gmail.com 写道：
char * s =''\'''
如果用分号终止上述语句，则它将指针s初始化为空指针。在这里，''\'''只是
另一种写作方式0.
请注意，它不会初始化s指向空字符串。要做
，你应该写：
const char * s ="" ;;
我使用const来提醒你字符串文字是不可写的。
如果你真的想要一个恰好包含
空字符串的可写缓冲区：
char s [100] ="" ;;
您必须指定缓冲区的大小。一个100字符缓冲区可以保存一个长达99个字符的
字符串，最后允许一个字节用于null
字符。
if（* s ==''\'''）{
这里你取消引用一个空指针。这可能会导致您的
程序崩溃。无论如何，它是未定义的行为。
printf（" yes"）;
}
上述代码是否可以正常运行？
No.
-
Simon。
char * s = ''\0''
if(*s==''\0''){
printf("yes");
}
Should the above code work without error?
cr*******@gmail.com said:
char * s = ''\0''
if(*s==''\0''){
printf("yes");
}
Should the above code work without error?No, it won''t even compile. Given that it''s just a code fragment, we''ll
overlook the lack of "furniture". But the lack of a semicolon on the first
line is not promising, is it? Also, although in this particular case the
value ''\0'' will actually /work/ as an initialisation for a pointer (because
it is simply an int with the value 0), "" would be more appropriate. Also,
you''ll want to make s const, to guard against accidentally writing through
it. And you will want to ensure that your text output stream is properly
terminated with a newline character:
#include <stdio.h>
int main(void)
{
const char *s = "";
if(*s == ''\0'')
{
printf("yes\n");
}
return 0;
}
--
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)
char * s = ''\0''
if(*s==''\0''){
printf("yes");
}
Okay sorry, forgot the semicolon. I''m sorry if it hurts you to see
plain code that''s not precise in adding an ending ''\n'' and main() and
what not--I was asking more conceptually rather than for actual use of
that code. I was looking through the framework code given to me for my
project and I believe I discovered a bug so I wanted to ask a
simplified question rather than give the whole piece.
cr*******@gmail.com wrote:char * s = ''\0''If you terminate the above statement with a semicolon, then it
initialises the pointer s to be a null pointer. Here, ''\0'' is just
another way of writing 0.
Note that it does not initialise s to point to an empty string. To do
that, you should write:
const char *s = "";
I made it const to remind you that string literals are not writable.
If you actually want a writable buffer that just happens to contain an
empty string:
char s[100] = "";
You have to specify a size for the buffer. A 100 char buffer can hold a
string up to 99 characters long, allowing one byte for the null
character at the end.
if(*s==''\0''){Here you are dereferencing a null pointer. This is likely to cause your
program to crash. At any rate, it is undefined behaviour.
printf("yes");
}
Should the above code work without error?No.
--
Simon.
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