如何使用getopt_long来解析多个参数？

Question

#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; 
}

期望的输出:

./a.out --stuff someArg1 someArg2

Hello
You hit stuff
Your agument(s): someArg1 someArg2
GoodBye!

Answer 1

当处理完所有选项时,getopt返回-1 .在--stuff被公认为是需要一个参数,在这种情况下的一个选项someArg1.该someArg2ARG不启动-或--,所以它不是一个选项.默认情况下,这将被置换到最后argv.之后的getopt返回-1,所有非选项ARGS将是argv来自optind于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-在开头添加单个optstring,getopt则返回1(不是'1')并指向optarg非选项参数:

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;
    }
}