使用printf打印字符串时遇到问题？ [英] Having problem on printing a string using printf?
问题描述
下面是我的代码，它没有printf。但是我通过使用for
语句并使用putchar到显示器，它在那里... ...
我如何才能使用printf？如果问题最后需要一个null
我该怎么办呢？
我在上一次记忆之后试过pos =''\'''也不起作用。
=================================== =============== ===========
void ERR（char * err_msg）{
char tmp [1024]，* pos;
unsigned int len，i，s;
memset（tmp，' '\0''，1024）;
len = strlen（err_msg）;
s = ntohl（len）;
pos = tmp;
memcpy（pos，& s，sizeof（s））;
pos + = sizeof（s）;
memcpy（pos，err_msg，len）;
printf（"％s \ n"，tmp）;
}
jt写道：这是我的代码如下，它不打印。但是我通过使用for
语句并使用putchar到显示器，它在那里... ...
如何让它使用printf？如果问题最后需要一个空的
我怎么能这样做呢？
我在上一次memcpy之后尝试了pos =''\'''，这也不起作用。< br => =============================================== === ===========
void ERR（char * err_msg）{
char tmp [1024]，* pos;
unsigned int len，i，s;
memset（tmp，''\''，1024）;
可能是浪费。 YMMV。
len = strlen（err_msg）;
s = ntohl（len）;
ntohl不是一个标准的C函数（这使得它在这里偏离主题）;它
甚至不是一个标准的POSIX功能（这使得它甚至可以在更多的地方使用它。）它存在的唯一功能通常是将网络* long *的字节顺序改为主机* long *。出于某种原因
您已选择将此转换的结果存储到int大小的
对象中。这不是一件好事。如果你使用这个非标准的
函数与你的问题有关，请在新闻组中询问 - 而不是
comp.lang.c - 它在哪里局部。
pos = tmp;
memcpy（pos，& s，sizeof（s））;
pos + = sizeof（s）;
memcpy（pos ，ERR_MSG，LEN）;
为什么不strcpy？哦，好吧......
printf（"％s \ n"，tmp）;
这是一个字节重新排序长的效果，存储在一个字符串的第一个字节中，然后放入字符串的第一个字节中是。如果这些字节中的任何
为0，那么你就死定了。试着看看
printf（"％s \ n"，pos）;
的确如此。 pos应该指向字符串的开头。
}
jt写道：这里是我的代码，它不是printf。但是我通过使用for
语句并使用putchar到显示器，它在那里... ...
如何让它使用printf？如果问题最后需要一个空的
我怎么能这样做呢？
我在上一次memcpy之后尝试了pos =''\'''，这也不起作用。< br => =============================================== === ===========
void ERR（char * err_msg）{
char tmp [1024]，* pos;
unsigned int len，i，s;
len更好地声明为size_t，因为它用于存储strlen（）的返回值
，即size_t。
memset（tmp，''\''，1024）;
或者，您可以声明tmp：
char tmp [1024] = {0};
len = strlen（err_msg）;
s = ntohl（len）;
pos = tmp;
memcpy（pos，& s，sizeof（s））;
pos + = sizeof （s）;
memcpy（pos，err_msg，len）;
您应该检查len的值。它不应该超过tmp的大小 -
sizeof s。
上面说你试图通过把pos =''\'''来解决这个问题。那个
会导致memcpy（）无用，如果你只是想稍后printf（）存储在tmp中的
字符串。
printf （％s \ n，tmp;
为了使其表现良好，你应该确保tmp [sizeof tmp
- 1] ==''\'''。
-
ishs
Martin Ambuhl写道（现在声称）：jt写道：
....void ERR（char * err_msg）{
char tmp [1024]，* pos;
unsigned int len，i，s;
memset（tmp，''\\ \\ 0''，1024）;
len = strlen（err_msg）;
s = ntohl（len）;
pos = tmp;
memcpy（pos，& s， sizeof（s））;
pos + = sizeof（s）;
memcpy（pos，err_msg，len）;
printf（"％s\ n"，tmp）;
它在空中是什么效果的字节重新排序长存储在一个
int然后放置进入字符串的第一个字节可能是。如果这些字节中的任何一个是0，那么你就死定了。试着看看
printf（"％s \ n"，pos）;
是什么。 pos应该指向字符串的开头。
[澄清]
除非err_msg非常长（长到足以放一个非-zero进入
* len的每个*字节以及s），你确定*有一个零字节
，其中包含tmp的第一个sizeof字节。这可以保证你的
字符串在存储的err_msg之前结束，并且除了
之外什么都没有看到，可能代表s的前导非零字节是什么乱码
as。但是err_msg是从pos开始存储的，它在*之后是*任意零
字节，所以如果你想看到
$ b，则放（pos）而不是puts（tmp） $ b err_msg。
}
Here is my code below, it doesn''t printf. But I step it thru using the for
statement and use putchar to the display, its there...
How can I get it to use printf? If the problem is needing a null at the end
and how can I do this?
I tried pos=''\0'' after my last memcpy and that doesn''t work either.
================================================== ===========
void ERR(char * err_msg){
char tmp[1024],*pos;
unsigned int len,i,s;
memset(tmp,''\0'',1024);
len=strlen(err_msg);
s = ntohl(len);
pos=tmp;
memcpy(pos,&s,sizeof(s));
pos+=sizeof(s);
memcpy(pos,err_msg,len);
printf("%s\n",tmp);
}
jt wrote:Here is my code below, it doesn''t printf. But I step it thru using the for
statement and use putchar to the display, its there...
How can I get it to use printf? If the problem is needing a null at the end
and how can I do this?
I tried pos=''\0'' after my last memcpy and that doesn''t work either.
================================================== ===========
void ERR(char * err_msg){
char tmp[1024],*pos;
unsigned int len,i,s;
memset(tmp,''\0'',1024); probably a waste. YMMV.
len=strlen(err_msg);
s = ntohl(len);
ntohl is not a standard C function (which makes it off-topic here); it
isn''t even a standard POSIX function (which makes it off-topic in even
more places). Where it exists, its sole function is usually to change
the byte order of a network *long* to a host *long*. For some reason
you have chose to store the result of this conversion into an int-sized
object. This cannot be a good thing. If your use of this non-standard
function has anything to do with your problem, ask in a newsgroup -- not
comp.lang.c -- where it is topical.
pos=tmp;
memcpy(pos,&s,sizeof(s));
pos+=sizeof(s);
memcpy(pos,err_msg,len); Why not strcpy? Oh, well...
printf("%s\n",tmp);
It is up in the air what the effect of a byte-reordered long stored in a
int and then placed into the first bytes of a string might be. If any
of those bytes are 0s, you''re dead. Try seeing what
printf("%s\n", pos);
does. pos should point to the beginning of the string.
}
jt wrote:Here is my code below, it doesn''t printf. But I step it thru using the for
statement and use putchar to the display, its there...
How can I get it to use printf? If the problem is needing a null at the end
and how can I do this?
I tried pos=''\0'' after my last memcpy and that doesn''t work either.
================================================== ===========
void ERR(char * err_msg){
char tmp[1024],*pos;
unsigned int len,i,s;
len is better declared as size_t as it''s used to store the return value
of strlen(), which is size_t.
memset(tmp,''\0'',1024);
Alternatively, you could declare tmp:
char tmp[1024] = {0};
len=strlen(err_msg);
s = ntohl(len);
pos=tmp;
memcpy(pos,&s,sizeof(s));
pos+=sizeof(s);
memcpy(pos,err_msg,len);
You should check the value of len. It shouldn''t exceed sizeof tmp -
sizeof s.
Above you said that you tried to solve this by putting pos = ''\0''. That
would cause the memcpy() to be useless if you just want to printf() the
string stored in tmp later.
printf("%s\n",tmp);
In order for this to behave well, you should ensure that tmp[sizeof tmp
- 1] == ''\0''.
--
ishs
Martin Ambuhl wrote (and now claifies):jt wrote: ....void ERR(char * err_msg){
char tmp[1024],*pos;
unsigned int len,i,s;
memset(tmp,''\0'',1024);
len=strlen(err_msg);
s = ntohl(len);
pos=tmp;
memcpy(pos,&s,sizeof(s));
pos+=sizeof(s);
memcpy(pos,err_msg,len);
printf("%s\n",tmp);
It is up in the air what the effect of a byte-reordered long stored in a
int and then placed into the first bytes of a string might be. If any
of those bytes are 0s, you''re dead. Try seeing what
printf("%s\n", pos);
does. pos should point to the beginning of the string.
[Clarification]
Unless err_msg is very very long (long enough to put an non-zero into
*every* byte of len and so of s), you are *sure* to have a zero byte
withing the first sizeof s bytes of tmp. This guarantees that your
string ends before the stored err_msg and that you see nothing but
whatever gibberish the leading non-zero bytes of s might be represented
as. But err_msg is stored starting at pos, which is *after* any zero
bytes in s, so puts(pos) rather than puts(tmp) if you want to see the
err_msg.
}
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