C++, command line, parameters

Question

I am starting my c++ program from command line:

program input_file1 input_file2 output_file

where

int main( int argc, char *argv[] )
{
    short indicator= 3;
    char input_file1[4096], input_file2[4096], output_file[4096];
    char *p_file = NULL;
    while ( --argc > 0 ) {
        switch (--indicator) {
          case 2:
            p_file = output_file;
            break;
          case 1:
            p_file = input_file2;
            break;
          case 0:
            p_file = input_file1;
            break;
        }

        for (char *argument = argv[argc]; ; ++argument) {
            if (*argument == '\0')
                break;
            else
                *p_file++ = *argument;
        }

        *p_file = '\0'; 
    }

    std::cout << input_file1 << '\n';
    std::cout << input_file2 << '\n';
    std::cout << output_file << '\n';
}

But with the real arguments

program D:\\data\\file1.txt D:\\data\\file2.txt D:\\data\\file3.txt

in names of the files only the first letter D is stored...

Output:
D
D
D

Thanks for your help...

Answer 1

Ok, so here is the short version:

int main(int argc, char *argv[]) {
    if (argc != 2) {
        std::cout << "This program requires 1 argument!" << std::endl;
        return 1;
    }
    std::string input_file(argv[1]);
    std::cout << input_file << std::endl;
}

You should be able to take it from here.

Answer 2

This is a C problem, not a C++ one, but as it is tagged C++, i will suggest a C++ solution for your problem :

int main( int argc, char *argv[] ) {
     std::vector<std::string> args(argv+1, argv+argc);
     std::cout << args[0] << '\n' << args[1] << '\n' << args[2] << std::endl;
}

UPDATE using iterators on argv to fill the vector args (thanks Space_C0wb0y)

Answer 3

Rather than copying the arguments, just set the file names to point to the appropriate entry in argv.

int main(int argc, char *argv[]) {
    char *input_file1, *input_file2, *output_file; 

    if (4 > argc)
    {
        std::cout << "Not enough parameters\n";
        return -1;
    }
    input_file1 = argv[1];
    input_file2 = argv[2];
    output_file = argv[3];

    std::cout << input_file1 << '\n';
    std::cout << input_file2 << '\n';
    std::cout << output_file << '\n'; 
}

the contents of argv will exist for as long as the program is running.

Answer 4

*p_file ++ = * argument; 

This assigns the first character of arguement to the first character in p_file.

You need to use strcpy, or declare some std::strings instead of arrays of char

Answer 5

Your loop is all wrong. You are looping through characters, not the parameter array. You can do that like this:

for(auto arg = argv + 1; *arg; ++arg)
{
    std::cout << (*arg) << '\n'; // *arg is a char*
}

This works because the arguments (if any) start at argv + 1 and because they are null terminated so that *arg is nullptr (converts to false) at the end of the argument array.

Your arguments are obtained by dereferencing arg using *arg which is a char*.

So you can do:

if(!std::strcmp(*arg, "--help"))
    print_help_info();