Re: oaklisp: contains 500kB binary in source
On Sun, Jun 13, 2004 at 04:17:29PM +0100, Marco Franzen wrote:
> It may not be a legal issue, but I think it is more than merely
> technical. It does touch the freeness question.
> It is about trust that the source provided is actually the true and full
> source for the given binary. This is not proven just because the source
> compiled with the binary reproduces the binary (even if it does).
> To understand what I mean, you may want to read Ken Thompson's old
> article on how to hide a Trojan Horse in a compiler without it being
> present in its "source" at all - just provided you bootstrap it with a
> given binary that already contains the Trojan Horse.
> Now just consider the *possibility* that the upstream bootstrap binary
> *might* contain such a self-reproducing Trojan Horse.
> One important aspect of Free Software is that you have (at least in
> principle) full control over what your computer does. It is your
> computer, not the s/w manufacturer's.
> Unless/until a clean and free bootstrap from scratch is available, it
> seems to me the source (with the bootstrap binary removed) may be
> contrib, but the binary could only be non-free.
> Unless/until it can be proved that the binary's behaviour is acurately
> described by its (alleged) source, it is unclear whether its (true)
> source is provided or missing. Erring on the side of caution, it would
> need to be ruled non-free.
> The source (with the bootstrap binary removed) could therefore be at
> most contrib.
How is this different from glibc?
Ok, I'm told it's possible to build glibc under bsd's libc, but
are we doing that? If oaklisp's binary can be built under some other
lisp implementation, is that sufficient?
What does "bootstrap from scratch" mean? Is it more important for
oaklisp than glibc?