在 C++ 中使用 getopt 时打印默认参数

Question

static struct option long_options[] =
{
     {"r",   required_argument,  0, 'r'},
     {"help",        no_argument,        0, 'h'},
     {0, 0, 0, 0}
};


int option_index = 0;
char c;
while((c = getopt_long(argc, argv, "r:h", long_options, &option_index)) != -1)
{
    switch(c)
    {
        case 'r':
            break;
        case 'h':
            return EXIT_SUCCESS;
    }
}

如何让 h 成为默认参数，所以如果这个程序运行没有任何参数，它就好像运行有 -h?

Answer 1

也许可以试试这样：

static struct option long_options[] =
{
     {"r",    required_argument,  0, 'r'},
     {"help", no_argument,        0, 'h'},
     {0, 0, 0, 0}
};

int option_index = 0;
char c = getopt_long(argc, argv, "r:h", long_options, &option_index);
if (c == -1)
{
    // display help...
    return EXIT_SUCCESS;
}

do
{
    switch(c)
    {
        case 'r':
            break;

        case 'h':
        {
            // display help...
            return EXIT_SUCCESS;
        }
    }

    c = getopt_long(argc, argv, "r:h", long_options, &option_index);
}
while (c != -1);

或者这样：

static struct option long_options[] =
{
     {"r",    required_argument,  0, 'r'},
     {"help", no_argument,        0, 'h'},
     {0, 0, 0, 0}
};

int option_index = 0;
char c = getopt_long(argc, argv, "r:h", long_options, &option_index);
if (c == -1)
    c = 'h';

do
{
    switch(c)
    {
        case 'r':
            break;

        case 'h':
        {
            // display help...
            return EXIT_SUCCESS;
        }
    }

    c = getopt_long(argc, argv, "r:h", long_options, &option_index);
}
while (c != -1);

Answer 2

为什么不创建一个 printUsage 函数并执行类似的操作。

if (c == 0) {
    printUsage();
    exit(-1);
}

在 C++ 中使用 getopt 时打印默认参数

Print default argument when using getopt in C++

c++

arguments

command-line-arguments

getopt

getopt-long