C getopt 多个值 [英] C getopt multiple value

问题描述

我的论点是这样的

./a.out -i file1 file2 file3

我如何利用 getopt() 获取 3 个(或更多)输入文件?我正在做这样的事情:

How can I utilize getopt() to get 3 (or more) input files? I'm doing something like this:

while ((opt = getopt(argc, argv, "i:xyz.."))!= -1){
  case 'i':
     input = optarg; 
     break;
  ...
}

我只得到file1；如何获取file2、file3?

I get just the file1; how to get file2, file3?

推荐答案

如果你必须，你可以从 argv[optind] 开始，然后自己增加 optind.但是，我建议不要这样做，因为我认为这种语法形式很差.(你怎么知道你什么时候到达了列表的末尾?如果有人有一个以 - 作为第一个字符命名的文件怎么办?)

If you must, you could start at argv[optind] and increment optind yourself. However, I would recommend against this since I consider that syntax to be poor form. (How would you know when you've reached the end of the list? What if someone has a file named with a - as the first character?)

我认为最好将您的语法更改为:

I think that it would be better yet to change your syntax to either:

/a.out -i file1 -i file2 -i file3

或者将文件列表视为位置参数:

Or to treat the list of files as positional parameters:

/a.out file1 file2 file3

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