c - Getopt- 为参数传递字符串参数

Question

我有一个接受多个命令行参数的程序，所以我正在使用 getopt。我的一个参数接受一个字符串作为参数。无论如何要通过 getopt 函数获取该字符串，还是我必须通过 argv[] 数组获取它？ getopt 也可以读取像 -file 这样的参数吗？到目前为止我看到的所有参数都只有一个字符，例如 -a

编辑

根据下面的答案，我编写了一个使用 getopt_long() 的程序，但是当我使用字符参数而不是长参数时，switch 语句只识别参数。我不确定为什么会这样。在传递参数 -mf -file sample 时，我没有看到打印语句。

编辑

我尝试将命令参数作为 --file 输入，然后成功了。仅使用 -file 是不可能做到这一点的吗？

static struct option long_options[] =
{
    {"mf", required_argument, NULL, 'a'},
    {"md", required_argument, NULL, 'b'},
    {"mn", required_argument, NULL, 'c'},
    {"mw", required_argument, NULL, 'd'},
    {"lf", required_argument, NULL, 'e'},
    {"ld", required_argument, NULL, 'f'},
    {"ln", required_argument, NULL, 'g'},
    {"lw", required_argument, NULL, 'h'},
    {"rf", required_argument, NULL, 'i'},
    {"rd", required_argument, NULL, 'j'},
    {"rn", required_argument, NULL, 'k'},
    {"rw", required_argument, NULL, 'l'},
    {"df", required_argument, NULL, 'm'},
    {"dd", required_argument, NULL, 'n'},
    {"dn", required_argument, NULL, 'o'},
    {"dw", required_argument, NULL, 'p'},
    {"file", required_argument, NULL, 'q'},
    {NULL, 0, NULL, 0}
};
int ch=0;
while ((ch = getopt_long(argc, argv, "abcdefghijklmnopq:", long_options, NULL)) != -1)
{
    // check to see if a single character or long option came through
        switch (ch){
        case 'a':
            cout<<"title";
            break;
        case 'b':
            
            break;
        case 'c':
            
            break;
        case 'd':
            
            break;
        case 'e':
            
            break;
        case 'f':
            
            break;
        case 'g':
            
            break;
        case 'h':
            
            break;
        case 'i':
            
            break;
        case 'j':
            
            break;
        case 'k':
            
            break;
        case 'l':
            
            break;
        case 'm':
            
            break;
        case 'n':
            
            break;
        case 'o':
            
            break;
        case 'p':
            
            break;
        case 'q':
            cout<<"file";
            break;
        case '?':
            cout<<"wrong message"
            break;  
    }
}

Answer 1

阅读 man getopt http://linux.die.net/man/3/getopt

optstring is a string containing the legitimate option characters. Ifsuch a character is followed by a colon, the option requires anargument, so getopt() places a pointer to the following text in thesame argv-element, or the text of the following argv-element, inoptarg. Two colons mean an option takes an optional arg; if there istext in the current argv-element (i.e., in the same word as the optionname itself, for example, "-oarg"), then it is returned in optarg,otherwise optarg is set to zero.

示例代码:

#include <stdio.h>
#include <unistd.h>

int main (int argc, char *argv[])
{
  int opt;
  while ((opt = getopt (argc, argv, "i:o:")) != -1)
  {
    switch (opt)
    {
      case 'i':
                printf("Input file: \"%s\"\n", optarg);
                break;
      case 'o':
                printf("Output file: \"%s\"\n", optarg);
                break;
    }
  }
  return 0;
}

这里在 optstring 中是“i:o:”，字符串中每个字符后的冒号“:”告诉这些选项需要一个参数。您可以在 optarg 全局变量中找到作为字符串的参数。有关详细信息和更多示例，请参阅手册。

对于多于一个字符的选项开关，请参见长选项 getopt_long。查看手册中的示例。

EDIT 以响应单个“-”长选项:

来自手册页

getopt_long_only() is like getopt_long(), but '-' as well as "--" can indicate a long option. If an option that starts with '-'(not "--") doesn't match a long option, but does match a short option,it is parsed as a short option instead.

查看手册并尝试一下。