命令行处理库 - getopt的 [英] command line processing library - getopt

问题描述

有人可以帮助我的getopt函数？

Can someone help me with the getopt function?

当我做主要如下：

char *argv1[] = {"testexec","-?"};
char *argv2[] = {"testexec","-m","arg1"};
int  cOption;
/* test for -? */

setvbuf(stdout,(char*)NULL,_IONBF,0);
printf("\n argv1 ");
while (( cOption = getopt (2, argv1, "m:t:n:fs?")) != -1) {
    switch(cOption){
        case 'm':
            printf("\n -m Arg : %s \n",optarg);
            break;
        case '?':
            printf("\n -? Arg ");
            break;
        case 'n':
            printf("\n -n Arg : %s \n",optarg);
            break;
    }
}

printf("\n argv2 ");

while (( cOption = getopt (3, argv2, "m:t:n:fs?")) != -1) {
    switch(cOption){
        case 'm':
            printf("\n -m Arg : %s \n",optarg);
            break;
        case '?':
            printf("\n -? Arg : %s \n",optarg);
            break;
        case 'n':
            printf("\n -n Arg : %s \n",optarg);
            break;
    }
}

我上运行RHEL3此code，它使用旧的libc版本。我不知道哪一个是准确的。

I'm running this code on rhel3 which uses old libc version. I don't know which one to be exact.

现在，问题是getopt的不与argv2工作的第二次。
但是，如果我注释掉与argv1第一getopt的呼叫，它的工作原理。

Now the problem is getopt doesn't work the second time with argv2. But if I comment out the first getopt call with argv1 , it works.

谁能告诉我我在做什么错在这里？

Can someone tell me what am I doing wrong here?

推荐答案

argv1和2必须以0结尾：

argv1 and 2 must end in 0:

char* argv1[] = {"par1", "par2", 0};

的修改的：OK，我读了getopt的手册页，我发现这一点：

Edit: OK, I read the getopt man page and I found this:

变量OPTIND是argv中要处理的下一个元素的索引。系统初始化该值
         1。调用者可以将其重置为1重启相同的argv扫描，扫描或一个新的参数向量时。

The variable optind is the index of the next element to be processed in argv. The system initializes this value to 1. The caller can reset it to 1 to restart scanning of the same argv, or when scanning a new argument vector.

所以，在getopt的在两个电话之间做出OPTIND = 1使它正常工作。

So, making optind=1 between the two calls at getopt makes it work as expected.

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