问题描述
#include <iostream>
#include <getopt.h>
#define no_argument 0
#define required_argument 1
#define optional_argument 2
int main(int argc, char * argv[])
{
std::cout << "Hello" << std::endl;
const struct option longopts[] =
{
{"version", no_argument, 0, 'v'},
{"help", no_argument, 0, 'h'},
{"stuff", required_argument, 0, 's'},
{0,0,0,0},
};
int index;
int iarg=0;
//turn off getopt error message
opterr=1;
while(iarg != -1)
{
iarg = getopt_long(argc, argv, "s:vh", longopts, &index);
switch (iarg)
{
case 'h':
std::cout << "You hit help" << std::endl;
break;
case 'v':
std::cout << "You hit version" << std::endl;
break;
case 's':
std::cout << "You hit stuff" << std::endl;
if(optarg)
std::cout << "Your argument(s): " << optarg << std::endl;
break;
}
}
std::cout << "GoodBye!" << std::endl;
return 0;
}
所需输出:
./a.out --stuff someArg1 someArg2
Hello
You hit stuff
Your agument(s): someArg1 someArg2
GoodBye!
推荐答案
getopt 当所有选项参数已处理完毕。 - stuff
被识别为接受参数的选项,在这种情况下 someArg1
。 someArg2
arg不以 -
或 -
,所以它不是一个选项。默认情况下,这将被置换到 argv
的结尾。在 getopt 返回-1之后,所有非选项参数将位于 argv
中 optind
到 argc-1
:
getopt returns -1 when all option args have been processed. The --stuff
is recognized as an option that takes an argument, in this case someArg1
. The someArg2
arg does not start with -
or --
, so it is not an option. By default, this will be permuted to the end of argv
. After getopt returns -1, all non-option args will be in argv
from optind
to argc-1
:
while (iarg != -1) {
iarg = getopt_long(argc, argv, "s:vh", longopts, &index);
// ...
}
for (int i = optind; i < argc; i++) {
cout << "non-option arg: " << argv[i] << std::endl;
}
如果添加单个 -
到 optstring
开始, getopt
将返回1(不是'1'), $ c> optarg 到非选项参数:
If you add a single -
to the start of optstring
, getopt
will return 1 (not '1') and point optarg
to the non-option parameter:
while (iarg != -1) {
iarg = getopt_long(argc, argv, "-s:vh", longopts, &index);
switch (iarg)
{
// ...
case 1:
std::cout << "You hit a non-option arg:" << optarg << std::endl;
break;
}
}
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