How to call execl() in C with the proper arguments?

i have vlc (program to reproduce videos) if i type in a shell:

/home/vlc "/home/my movies/the movie i want to see.mkv"

it opens up an reproduces the movie.

however, when I run the following program:

#include <unistd.h> int main(void) { execl("/home/vlc", "/home/my movies/the movie i want to see.mkv",NULL); return 0; } 

vlc opens up but doesn't reproduce anything. How can I solve this?

Things I tried:

I guessed

execl("/home/vlc", "/home/my movies/the movie i want to see.mkv",NULL); 

was equivalent to typing in the shell:

/home/vlc /home/my movies/the movie i want to see.mkv 

which doesn't work, so i tried

 execl("/home/vlc", "\"/home/my movies/the movie i want to see.mkv\"",NULL); 

and vlc opens up but doesn't reproduce either.

Instead of writing NULL at the end I tried 0, (char*) 0, 1 .... not helpful. Help!!!!

2 Answers

execl("/home/vlc", "/home/vlc", "/home/my movies/the movie i want to see.mkv", (char*) NULL); 

You need to specify all arguments, included argv[0] which isn't taken from the executable.

Also make sure the final NULL gets cast to char*.

Details are here:

If you need just to execute your VLC playback process and only give control back to your application process when it is done and nothing more complex, then i suppose you can use just:

system("The same thing you type into console");