[magicalhippo]

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/e...

[skobes]

> Windows does not allow you to do that, AFAIK.

It does though, using the lpCommandLine parameter to CreateProcess as I said.

CreateProcess("main.exe", "foobar", ...)

argv[0] is "foobar"

[magicalhippo]

I stand corrected. Been ages since I used Win32 API a lot, and I realized I can't recall using both of those arguments when calling CreateProcess.