Thread: Passing Arguments in windows

Look up what's in this FAQ page, oh and this

Here check out this code, it checks if it got two arguments on the commandline(other than the path to the executable) and if it did it displays them. From this you should be able to utilize the arguments.

Code:

#include <iostream>

using namespace std;

int main(int argc, char* argv[])
{
    if(argc==3)
    cout<<endl<<argv[1]<<endl<<argv[2]<<endl;
    cin.get();
    return 0;
}

The program always has one argument so argc is always one at least and argv[0] is always the path to the executable. After that are the arguments you entered.

Is argv[0] garrenteed to be the path to the executable?

As far as I know, yes. My understanding is that the path to the executable is passed as the first argument because it needs to use that path to execute it. Please correct me if I'm wrong.

No. It was discussed on the C board.

Originally Posted by Grumpy

Not quite. It either represents the program name or it is an empty string.

There are all sorts of possibilities in that, depending on how the word "represents" is interpreted. For example argv[0] can be any of these;

1) the unadorned name of the executable (eg "program")

2) the name of the executable file (eg "program.exe" under windows)

3) the fully qualified path of the executable file (eg "c:/directory/subdirectory/program.exe" or "C:\directory\subdirectory\program.exe" under windows, or drive:[directory.path]program.exe on an old VAX/VMS system).

4) an alias for the executable (eg several unix shells support aliases, and could be passed as argv[0] by the shell, rather than the actual name of the executable)

5) the name of something that is linked to the executable (eg symbolic links under unix: the name of the link can be passed as argv[0]).

That is what I thought, but I wasn't sure if the rules of C argument passing applied in C++ (since their standards have branched and some things arent the same.)