c++ - 使用 getopt_long (C++) 我如何编写长选项和短选项以同时需要参数？

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, "svh", 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;
        break;
    }
  }

  std::cout << "GoodBye!" << std::endl;

  return 0; 
}

输出:

./a.out -s
Hello
You hit stuff
GoodBye!

输出:

./a.out --stuff
Hello
./a.out: option `--stuff' requires an argument
GoodBye!

需要解决的冲突 -s 和 --s 都应该说:./a.out: option `--stuff' requires an argument 当使用而不继续时命令后的参数。但只有 --stuff 可以吗？有谁知道我在这里缺少什么？

期望的结果:

./a.out -s
     Hello
     ./a.out: option `--stuff' requires an argument
      GoodBye!

./a.out --stuff
     Hello
     ./a.out: option `--stuff' requires an argument
     GoodBye!

Answer 1

您的 getopt_long 调用为短选项指定了“svh”，而它应该是“s:vh”。冒号告诉 getopt 需要“s”的参数。