Command line arguments: argv

Question

I would like to know why argv array can hold multi-character elements in the following code:

#include <iostream>

int main(int argc, char *argv[])
{
  for (int nArg = 0; nArg < argc; nArg++)
  {
    cout << nArg << " " << argv[nArg] << endl;
  }

  return 0;
}

The code just prints out all arguments passed to the program.

When I run the program with arguments foo, and bar, argv[1] is foo and argv[2] is bar.

But argv is an array of char. And foo is not char, but a string. How can argv[0] be foo?

Answer 1

argv is actually a pointer, not an array. In both C and C++, a function parameter cannot be an array. If it looks like an array, it's quietly "adjusted" at compile time to be a pointer to the array's element type.

This:

int main(int argc, char *argv[])

really means this:

int main(int argc, char **argv)

When your program is invoked, argv will be initialized to point to a char* object. That object is the initial element of an array of char* pointers, each of which points to (the initial element of) a string -- except that the final element, argv[argc] contains a null pointer.

argv[0] is a pointer to a string that represents the name of the program. If the program is invoked with two arguments, argv[1] and argv[2] will point to those arguments.

Finally, if you print a char* value using std::cout << ..., it will print, not the pointer value itself, but the value of the string that it points to.

When I run the program with arguments foo, and bar, argv[0] is foo and argv[1] is bar.

Are you sure about that? (Update: That was a typo in the question, now corrected.) If your program is named "my_program", the output of my_program foo bar should be something similar to:

0 my_program
1 foo
2 bar

Answer 2

But argv is an array of char.

Correction: it's an array of char *. "foo" is a char *, so this works.

An array of char is char argv[].
An array of char * is char *argv[]. That's what you have.

Answer 3

argv[0] is foo and argv[1] is bar ...

You should note that argv[0] always receives the actual program name called from the shell, so your command line arguments start from argv[1].

You're just off by one.

Answer 4

But argv is an array of char.

It's not. It's an array of char pointers, and those pointers point to (zero-terminated) char arrays, i.e. C-style strings.

Technically, argv is not an array, but a pointer to an array (the array containing the char pointers). See Keith Thompson's comment on John Kugelman's answer for more details.