Parsing command line arguments without specifying the number of arguments C++

The signature

int main(char* argv[])

is not supported by any C++ compiler I know of. Admittedly very few nowadays. But still.

The signature

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

is standard and works well for pure ASCII arguments.

It also works well for general Unicode arguments in Unix-land, when they're encoded as UTF-8.

Anyway, it tells you the number of arguments.

argc tells you how many arguments the program received. So argv[0] refers to the name of the program itself, argv[1] to the first argument you passed on the command line, argv[2] to the next, and so on up to argv[argc-1], which is the last one. That's followed by argv[argc], which is a null pointer.

So, when you run the program, you don't have to specify the number of arguments--you just pass whatever number of arguments you want to. Some code in the library then goes through, splits what you pass up into individual arguments, and tells the program how many it found.

IOW, when you run:

testProgram arg1 arg2 arg3

your program is going to receive:

argv[0] = testProgram
argv[1] = arg1
argv[2] = arg2
argv[3] = arg3
argv[4] = null pointer
argc = 4

Note: the exact content of argv[0] is open to variation--it might be exactly what you typed to invoke the command, or it might be a full path to the command, or it might (especially on ancient systems) be an empty string. On most modern systems, it'll normally contain something like the file name though.

You can ignore the argc argument and process the argv[] elements until you get to the one that's NULL since the argv array is specified to have a sentinel element that's NULL. I'm not sure what the ultimate point of the question is though.

#include <stdio.h>

int main(int argc, char** argv) 
{
    while (*argv) {
        puts(*argv);
        ++argv;
    }
    return 0;
}