mjhjmtu

I am trying to use execvp() to run less however I keep running into the same error saying that:

Missing filename ("less --help" for help)

I'm assuming I am trying to input the file completely wrong. Could anyone give me some guidance? Here is my code line trying to implement it:

// args[0] == "tempFile" which is in my directory
execvp("less", args)

Just as when you execute a shell script on the command line script

sh -c 'script body here' arg0 arg1 arg2

the arg0 argument gets placed in $0 (this is usually the name of the process). The first command line argument available in $1 is arg1.

In your case, use args[0] = "less" and args[1] = "tempFile" in your C code. args[3] should be a null pointer.

argv[0] is meant to be the name you give to the command. The command you execute uses that to know how it was invoked. Typically, you'd want to use something like less here:

argv[0] = "less";
argv[1] = "filename";
argv[2] = NULL;
execvp("less", argv);

Argument 0, taken from argv[0], is conventionally the command name. Typically it's the same string that you pass as the first argument to execvp (that's what shells do).

Argument 1, from argv[1], is the first “real” argument. So pass a 3-element array containing: char *args = "less", "tempFile", NULL

Most languages follow the same argument numbering. For example, if you invoke a shell script, it sees what you pass as argv[0] as its $0, what you pass as argv[1] as $1, etc. Perl is a notable exception: in a Perl script, argv[0] is $0, argv[1] is $ARGV[0], argv[2] is $ARGV[1], etc.