解析选项的方式与传递给main的方式相同 [英] Parsing options in the same way they are passed to main
问题描述
如果main的原型为：
int main（int argc，char * argv []）;
你最终会得到一堆* argv中的参数和argc中的数字。现在
我想要做的就是在字符串上模拟相同的动作。比方说我有：
char * command =" ./ blah --arg1 --arg2 123 -x --arg2 = w00t"
如何实现与main（）相同的效果？
Jeff
If main is prototyped as:
int main(int argc, char *argv[]);
You will end up with a bunch of arguments in *argv, and the number in argc. Now
what I want to do is emulate that same action on a string. Say for example I have:
char *command = "./blah --arg1 --arg2 123 -x --arg2=w00t"
How do I acheive the same effect as in main()?
Jeff
推荐答案
Jeff Rodriguez<是ne ******** @ gurugeek.EXAMPLENOSPAM.com>写道：
Jeff Rodriguez <ne********@gurugeek.EXAMPLENOSPAM.com> writes:
如果main的原型为：
int main（int argc，char * argv []）;
你最终会得到* argv中的一堆参数，以及
argc中的数字。现在我想做的是在
字符串上模拟相同的动作。比方说我有：
char * command =" ./ blah --arg1 --arg2 123 -x --arg2 = w00t"
我如何实现与main（）相同的效果？
If main is prototyped as:
int main(int argc, char *argv[]);
You will end up with a bunch of arguments in *argv, and the number in
argc. Now what I want to do is emulate that same action on a
string. Say for example I have:
char *command = "./blah --arg1 --arg2 123 -x --arg2=w00t"
How do I acheive the same effect as in main()?
这取决于操作系统，shell和其他东西。如果
你要做的就是将字符串拆分为
空格中的单词，我建议你只需编写代码即可。它不是太难了。
-
鉴于计算能力随时间呈指数增长，
$带有指数或更好的O符号的b $ b算法
实际上是一个大的常数线性。
--Mike Lee
It depends on the operating system, shell, and other things. If
all you want to do is to break apart the string into words at
white space, I suggest you just write code to do it. It''s not
too hard.
--
"Given that computing power increases exponentially with time,
algorithms with exponential or better O-notations
are actually linear with a large constant."
--Mike Lee
Ben Pfaff写道：
Ben Pfaff wrote:
这取决于操作系统，shell和其他东西。如果您想要做的就是将字符串拆分为白色空间中的单词，我建议您只需编写代码即可。这不太难。
It depends on the operating system, shell, and other things. If
all you want to do is to break apart the string into words at
white space, I suggest you just write code to do it. It''s not
too hard.
确实:(原始代码）
#include< stdio.h>
#include< string.h>
#include< unistd.h>
#include< stdlib.h>
int main（int argc，char * argv []）
{
int i;
char * string = NULL;
char * str = NULL;
int size = 0;
for（i = 1 ; i< argc; i ++）
{
if（strlen（argv [i]）> size）
{
if（（string = realloc（string，（size + strlen（argv [i]））* 2））== NULL）
{
perror（无法分配内存）;
退出（1）;
}
}
string = strcat（string，argv [i]）;
string = strcat（string，""）;
}
printf（"％s％s \ n"，argv [0]，string）;
str = strtok（string，" \ t \ n"）;
printf（&quo t;％s \ n"，str）;
while（（str = strtok（NULL，" \ t \ n"））= = NULL）
{
printf（"％s \ n"，str）;
}
返回0;
}
然而你遇到的问题包括：
../a.out --search =" Hello World！"
我宁愿不必重新发明轮子，因为main（）已经
实现它！必须/某些/方式它是内置功能！
Jeff
Indeed: (crude code)
#include <stdio.h>
#include <string.h>
#include <unistd.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
int i;
char *string = NULL;
char *str = NULL;
int size = 0;
for ( i = 1; i < argc; i++ )
{
if ( strlen(argv[i]) > size )
{
if ( (string = realloc(string, (size + strlen(argv[i])) * 2)) == NULL )
{
perror("Could not allocate memory");
exit(1);
}
}
string = strcat(string, argv[i]);
string = strcat(string, " ");
}
printf("%s %s\n", argv[0], string);
str = strtok(string, " \t\n");
printf("%s\n", str);
while ( (str = strtok(NULL, " \t\n")) != NULL )
{
printf("%s\n", str);
}
return 0;
}
However you run into problems like:
../a.out --search="Hello World!"
I would prefer not to have to reinvent the wheel on this since main() already
implements it! There must be /some/ way that it''s a built-in function!
Jeff
12月12日星期五2003 22:02:24 -0700，Jeff Rodriguez
< ne ******** @ gurugeek.EXAMPLENOSPAM.com>在comp.lang.c中写道：
On Fri, 12 Dec 2003 22:02:24 -0700, Jeff Rodriguez
<ne********@gurugeek.EXAMPLENOSPAM.com> wrote in comp.lang.c:
Ben Pfaff写道：
Ben Pfaff wrote:
这取决于操作系统，shell和其他东西。如果您想要做的就是将字符串拆分为白色空间中的单词，我建议您只需编写代码即可。它确实不太难。
确实:(原始代码）
#include< stdio.h>
#include< string.h>
#include< unistd.h>
It depends on the operating system, shell, and other things. If
all you want to do is to break apart the string into words at
white space, I suggest you just write code to do it. It''s not
too hard.
Indeed: (crude code)
#include <stdio.h>
#include <string.h>
#include <unistd.h>
上面的标题在您的程序中都是非标准和不需要的。
#include< stdlib.h>
int main（int argc，char * argv []）
{
int i;
char * string = NULL;
char * str = NULL;
int size = 0;
for（i = 1; i< argc; i ++）
{
if（strlen（argv） [i]）> size）
{
if（（string = realloc（string，（size + strlen（argv [i]））* 2））== NULL）
{
perror（无法分配内存）;
退出（1）;
}
string = strcat（string，argv [一世]）;
这很可能第一次崩溃，因为realloc返回的第一个块
是未初始化的，并且不知道它是否是
将包含一个终止''\ 0''，更不用说在第一个
元素中。
string = strcat（string，" ;"）;
}
printf（"％s％s \ n"，argv [0]，string）;
str = strtok （string，" \t\\\
"）;
printf（"％s \ n"，str）;
while（（str = strtok（NULL，" \ t） \ n"））！= NULL）
{
printf（"％s \ n"，str）;
}
返回0;
}
然而你遇到的问题包括：
./a.out --search =" Hello World！"
我宁愿不要重新发明轮子，因为main（）已经实现了它！必须/某些/方式它是内置功能！
杰夫
The header above is both non-standard and unneeded in your program.
#include <stdlib.h>
int main(int argc, char *argv[])
{
int i;
char *string = NULL;
char *str = NULL;
int size = 0;
for ( i = 1; i < argc; i++ )
{
if ( strlen(argv[i]) > size )
{
if ( (string = realloc(string, (size + strlen(argv[i])) * 2)) == NULL )
{
perror("Could not allocate memory");
exit(1);
}
}
string = strcat(string, argv[i]);
This is quite likely to crash the first time, since the first block
that realloc returns is uninitialized and there''s no telling if it
will contain a terminating ''\0'' at all, let alone in the first
element.
string = strcat(string, " ");
}
printf("%s %s\n", argv[0], string);
str = strtok(string, " \t\n");
printf("%s\n", str);
while ( (str = strtok(NULL, " \t\n")) != NULL )
{
printf("%s\n", str);
}
return 0;
}
However you run into problems like:
./a.out --search="Hello World!"
I would prefer not to have to reinvent the wheel on this since main() already
implements it! There must be /some/ way that it''s a built-in function!
Jeff
不，main（）还没有实现它。主机运行
系统或实现的启动代码都可以。这样做没有C
标准库函数。
如果你想要源代码，请试试Google。
-
Jack Klein
主页： http://JK-Technology.Com
常见问题解答
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c ++ http：//www.parashift .com / c ++ - faq-lite /
alt.comp.lang.learn.c-c ++ ftp://snurse-l.org/pub/acllc-c++/faq
No, main() does not already implement it. Either the host operating
system or the implementation''s start-up code does. There is no C
standard library function to do so.
If you want source code, try Google.
--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
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