如果我使用结构来承载argc和argv，如何将地址argv分配给结构内的变量？

Question

I have defined some structure above main as:

struct arguments
{
   int c;
   char *v[];
};

now I want to do this in main:

int main(int argc, char *argv[])
{
    arguments arg;

    arg.c = argc;
    arg.v = argv; /* error: incompatible types in assignment of 'char** to char* [0]' */
}

So, I do not fully understand what I am doing, but instinctively I remade the structure such that the line char *v[]; is instead char **v[]; however, this means when I pass by reference my structure arg , if I want to dereference and get the value of arg.v[0] , which will be the program name, I now have to do *arg.v[0] and this no longer works for *arg.v[1] .

I have a function such as:

void argument_reader(arguments &arg)
{
   cout << "Number of arguments: " << arg.c << endl << endl;
   cout << "Array\t\tValue\n";
   cout_bar(40);
   for (int i = 0; i < arg.c; i++)
   {
      cout << "argv[" << i << "]\t\t" << *arg.v[i] << endl;
   }
   return;
}

but when I call *arg.v[i] the program ends without printing the value of the second argument or any others for that matter.

So my question is, is there some way I can use the struct to pass between functions such that I can call arg.v[some_index < arg.c] inside another function by somehow making the line arg.v = argv in main work as I had intended (to store the address of argv in arg.v )

All the structure does is make it so that instead of having to pass argc and argv to a function which has a decleration like: void func(int &argc, char *argv[]); I can instead pass my new structure type with variable name arg to a function declared like void func(arguments &arg) and that is basically all I wanted the structure for. So if I cannot do this, then I can go back to the first method.

Answer 1

You need to write:

struct arguments
{
    int    c;
    char **v;
};

Now your main() will work. (And this sort of confusion is why I always use int main(int argc, char **argv) , but that's a personal quirk.)

If you were working in C99 or later, what you created with your structure is what is called a 'flexible array member' (FAM), an addition to C99 over C89 (and therefore over C++98). I didn't notice that you're working in C++! There are special rules about FAM in C, and you can't do the assignment as you did with a flexible array member.

However, the fix will work in C++ just as much as in C; it is just not necessarily correct that you're using an FAM. Are you using a C++ compiler that supports FAM as an extension?

In your output line, you have:

cout << "argv[" << i << "]\t\t" << *arg.v[i] << endl;

That would print the first character of the argument; drop the * and you'd get the first argument as a string:

cout << "argv[" << i << "]\t\t" << arg.v[i] << endl;

Answer 2

Change your struct to this:

struct arguments
{
   int c;
   char **v;
};

C-style array and pointer are very similar, and you can use above struct just like your version or the original argv parameter.

Indeed, often main function is declared like that instead of using []:

int main(int argc, char **argv)

Note that with this answer code v points to the original argv array of pointers to char, pointers are not copied. If you actually want a copy, you need to dynamically allocate array of char pointers and copy.

Answer 3

two things:

struct arguments
{
   int c;
   char *v[]; /*unknown length, hence structure is incomplete*/
};

You need a double pointer, which holds the address of the argv

    struct arguments {
       int c;
       char **v;
    };