C: execvp() and command line arguments

Question

So I'm writing a program where the arguments are as follows:

program start emacs file.c

or even

program wait

In essence, the first argument (argv[0]) is the program name, followed by user inputs.

Inside my code, I invoke execvp. Thing is, I'm not entirely sure I'm invoking the right arguments.

if (pid == 0) { 
                    execvp(argv[1], argv); //Line of interest
                    exit(1);
                    }

are argv[1] and argv the correct arguments to pass for the functionality described above? I looked at the man page and they make sense but might not be correct for this case. Thank you!

Answer 1

In your main, argv will be like this in the first example:

argv[0] = "program";
argv[1] = "start";
argv[2] = "emacs";
argv[3] = "file.c";
argv[4] = NULL;

In execv you want to execute the program "start" with args "emacs file.c", right?. Then the first parameter should be argv[1] - "start" and the second one an array with this strings: {"start", "emacs", "file.c", NULL}. If you use argv, you include the "program" string in argv[0].

You can create a new array and copy these parameters or use the address of argv[1] like this:

execvp(argv[1], &argv[1]); //Line of interest

Answer 2

The only thing that might be an issue is that argv[0] in argv passed to execvp won't match argv[1] (the first argument). Otherwise, it looks okay.

Imagine calling program cat file.txt. In your program, argv will be {"program", "cat", "file.txt", NULL}. Then, in cat, even though the binary called will be cat, argv will still be {"program", "cat", "file.txt", NULL}.

Since cat tries to open and read each argument as a file, the first file it'll try to open is cat (argv[1]), which isn't the desired behavior.

The simple solution is to use execvp(argv[1], argv+1) - this essentially shifts the argument array to the left by one element.

Answer 3

My understanding is that you want to take a specific action based on the second command-line argument (argv[1]). If the second argument is 'start', your program should start the executable named argv[2] with the arguments provided thereafter (right?). In this case, you should provide execvp with the executable name (argv[2]) [1] and a list of arguments, which by convention starts with the name of the executable (argv[2]).

execvp(argv[2], &argv[2]) would implement what we have described in the last paragraph (assuming this is what you intended to do).

[1] execvp expects 2 arguments as you know. The first is a filename; if the specified filename does not contain a slash character (/), execvp will do a lookup in the PATH environment variable (which contains a list of directories where executable files reside) to find the executable's fully-qualified name. The second argument is a list of command-line arguments that will be available to the program when it starts.