getopt的转变OPTARG [英] Getopt shift optarg

问题描述

我需要调用我的程序是这样的：

I need to call my program like this:

./program hello -r foo bar

我带你好出来的argv [1]，但我有超​​值的酒吧麻烦，也应该我改变R：别的东西。

I take hello out of argv[1], but i am having trouble with value bar, also should i change "r:" to something else?

while((c = getopt(argc, argv, "r:")) != -1){
   switch(i){
   ...
   case 'r':
     var_foo = optarg;
     //shell like argument shift here?
     var_bar = optarg;
     break;
...}

我知道我可以通过的argv做到这一点，但有一种用getopt的类似的方式做到这一点是在bash？

I know I could do this with passing through argv, but is there a way to do it with getopt similar way as in bash?

感谢。

推荐答案

栏不在眼里的选项参数 getopt的。相反，GNU getopt的的重新排列的位置参数，以便在处理结束时，你有的argv [3] 是你好和的argv [4] 是栏。基本上，你就大功告成了getopting的时候，你还有位置参数 [OPTIND，ARGC）来处理：

bar is not an option argument in the eyes of getopt. Rather, GNU getopt rearranges the positional arguments so that at the end of the processing, you have argv[3] being "hello" and argv[4] being "bar". Basically, when you're done getopting, you still have positional arguments [optind, argc) to process:

int main(int argc, char * argv[])
{
     {
         int c;
         while ((c = getopt(argc, argv, ...)) != -1) { /* ... */ }
     }

     for (int i = optind; i != argc; ++i)
     {
         // have positional argument argv[i]
     }
}

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