From:	"Böhme Uwe" <uwe AT hof DOT baynet DOT de>
Newsgroups:	comp.os.msdos.djgpp
Subject:	Re: please help {newbie}
Date:	Wed, 29 Jul 1998 12:57:40 +0200
Organization:	Bingo (Buergernetz Ingolstadt eV)
Lines:	24
Message-ID:	<35BF0024.B4BCDF87@hof.baynet.de>
References:	<6pmncl$pph$1 AT osprey DOT global DOT co DOT za>
NNTP-Posting-Host:	port29.hof.baynet.de
Mime-Version:	1.0
To:	djgpp AT delorie DOT com
DJ-Gateway:	from newsgroup comp.os.msdos.djgpp

> how do i read a file name to the program in the calling statement i.e.
> "c:\resolve test.txt"
>
> thanks
> heinrich

  If a program is called with command line arguments (or without) you'll
get two parameters passed to your main()

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

The argc argument keep the number of command line arguments (at least 1,
even if no argument is specified),
argv[0] is a pointer to a string wich is holding the path of the exe file
itself (i.e. "c:\resolve.exe").
In your example argc would be 2 and argvc[1] would point to "test.txt".
Additional arguments increase argc and the n'th argument can be read by
argv[ n - 1 ].

Good luck
Uwe

- Raw text -