Not the way I understand it. In the execv documentation[1], you pass the program name twice: int execv(const char *path, char *const argv[]); The argument pat

Not the way I understand it. In the execv documentation[1], you pass the program name twice: int execv(const char *path, char *const argv[]); The argument path points to a pathname that identifies the new process image file. The argument argv is an array of character pointers to null-terminated strings. [..] The value in argv[0] should point to a filename string that is associated with the process being started by one of the exec functions. Windows does not allow you to do that, AFAIK. [1]: https://pubs.opengroup.org/onlinepubs/9699919799/functions/execve.html https://pubs.opengroup.org/onlinepubs/9699919799/functions/e...