Incomplete argv/argc when using FTYPE and ASSOC

I'm trying to use a standard C main() program with FTYPE and ASSOC.

The main.c program is:

Code:

main(argc,argv) 
int argc; 
char **argv; 
{ 
int i; 


printf("argc = %d\n", argc); 


for(i=0; i<argc; i++) 
printf ("argv[%d] = %s\n", i, argv[i]); 

}

And I use in a DOS windows:

Code:

ASSOC .ma=MainScript 
FTYPE MainScript=main.exe %1 %*

If I call the program in this way:

Code:

testma.ma 1 2 3

I only get:

Code:

argc = 2 
argv[0] = main.exe 
argv[1] = testma.ma

The command line arguments ("1", "2" and "3") here are not visible in
argc/argv.

How do I retrieve these arguments in the C program?

Thank you in advance for any help.

The arguments aren't passed to your program. The problem lies elsewhere.
Your code is very poorly indented and might need some help:
http://cpwiki.sf.net/Indentation

Code:

main(argc,argv) 
int argc; 
char **argv;

Is incorrect syntax.
It should be

Code:

int main(int argc, char** argv)

Originally Posted by Elysia

The arguments aren't passed to your program. The problem lies elsewhere.

I know, that's my problem!

Using exactly the same FTYPE and ASSOC syntax with other commercial programs (like the Hamilton C-Shell) works, they do get the arguments.

So, I'm stumped. I just can't figure it out.

Your code is very poorly indented and might need some help:

That was just a quick sample to reproduce the problem I have with the real application.

Thanks.

Nevertheless, indentation is important and should never be left out. Even if it just a demonstration or sample code.