这是犹太人吗？ [英] Is this kosher?
问题描述
伙计，
我刚写了一个函数来找到一行的结尾：
char * eol（char * pt ）
{while（* pt！=''\'''&& * pt！=''\ n''&& * pt！=''\ 0''）//跑到线上
终结者角色
pt ++;
返回pt; //并返回指向它的指针
}
我这样使用它：
main（）/ /我知道，我知道！
{char text_line [260]，* t_pt;
/ * [获取文本路径到text_line] * /
t_pt = eol（text_line）;
* t_pt =''\'''; //这会敲掉一个''\ n''，它会停止chdir（）从
工作
chdir（text_line）;
}
这样工作正常，但仅仅是为了实验，我更换了两行
含有t_pt的
* eol（text_line）=''\ 0'';
....并且它有效。这是一个语法上的事情吗？它是便携式的吗？我会
感谢你对此的教育评论。
多亏一堆，
MikeC。< br $>
-
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Folks,
I just wrote a little function to find the end of a line:
char *eol(char *pt)
{ while(*pt != ''\r'' && *pt != ''\n'' && *pt != ''\0'') // run along to the line
terminator character
pt++;
return pt; // and return a pointer to it
}
I use it thus:
main() // I know, I know!
{ char text_line[260], *t_pt;
/* [ get a file path into text_line] */
t_pt = eol(text_line);
*t_pt = ''\0''; // this knocks off a ''\n'', which stops chdir() from
working
chdir(text_line);
}
This works fine, but just for the experiment, I replaced the two lines
containing t_pt with
*eol(text_line) = ''\0'';
.... and it worked. Is this a syntactical thing to do? Is it portable? I''d
appreciate your educated comment on this.
Thanks a heap,
MikeC.
--
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mike_best$ntlworld*com
$ = @
* = dot
推荐答案
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MikeC说：
< snip>
MikeC said:
<snip>
>
这样工作正常，但仅仅是为了实验，我更换了两行
包含t_pt with
* eol（text_line）=''\ 0'';
......它有效。这是一个语法上的事情吗？它是便携式的吗？
>
This works fine, but just for the experiment, I replaced the two lines
containing t_pt with
*eol(text_line) = ''\0'';
... and it worked. Is this a syntactical thing to do? Is it portable?
是的，没关系，只要你绝对100％确定eol（）返回的指针
是保证的是一个可写的角色。在你的
情况下，只要你只给eol（）提供可写字符串。
只给eol（）提供真正的以空字符结尾的字符串 - 一个char数组
不一定/不包含字符串：char foo [3] =" foo" ;;是一个没有的例子。如果你计划做你的* eol（）的事情，
不要给eol（）一个字符串文字来提供：
* eol （哦，deary deary me\ n）=''\'''; / *要求麻烦* /
char * p =" oh deary deary me\ n";
* eol（p）='' \0 ''; / *仍然要求同样的麻烦* /
char arr [] ="但这没关系\ n"）;
* eol（ arr）=''\'''; / *没有汗水* /
-
Richard Heathfield
Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk
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Yes, it''s fine, provided you are absolutely 100% certain that the pointer
that eol() returns is guaranteed to be to a writeable character. In your
case, it is, provided that you only give writeable strings to eol().
Only give genuine null-terminated strings to eol() - an array of char
doesn''t /necessarily/ contain a string: char foo[3] = "foo"; being an
example of one that does not. And if you plan to do your *eol() thing,
don''t give eol() a string literal to feed on:
*eol("oh deary deary me\n") = ''\0''; /* asking for trouble */
char *p = "oh deary deary me\n";
*eol(p) = ''\0''; /* still asking for the same trouble */
char arr[] = "but this is okay\n");
*eol(arr) = ''\0''; /* no sweat */
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
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