Question

execvp(argv[1], &argv[1])

What exactly is done with the second argument of execvp()?

Answer 1

The second argument should be a pointer to a NULL-terminated array of strings, which becomes the argv of the called process.

The first element of this array becomes the argv[0] of the callee, which is not necessarily the same as its path; e.g., you can call a process by its full path, but pass it its basename as argv[0]. Also, some programs behave differently based on their argv[0]. The famous example is that Unix shells behave as “login shells” when their argv[0] starts with a -, so a primitive login program could do

char *argv[2] = {"-sh", NULL};
execvp("/bin/sh", argv);