Messages in this thread | | | Date | Fri, 26 Mar 2004 09:50:09 +0100 | From | Stefan Smietanowski <> | Subject | Re: Binary-only firmware covered by the GPL? |
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Hi David.
> The firmware blob in question can be reasonably considered to be an > independent and separate work in itself. The GPL doesn't apply to it > when it is distributed as a SEPARATE work. But when you distribute it as > part of a whole which is a work based on other parts of the kernel, by > including it in the kernel source in such a manner, the distribution of > the whole must be on the terms of the GPL, whose permissions for other > licensees extend to the entire whole, and thus to each and every part. > > It's not the intent of the GPL to claim rights to firmware written > independently for such hardware; rather, the intent is to exercise the > right to control the distribution of _COLLECTIVE_ works based on the > indisputably GPL'd parts of the kernel. >
But the firmware comes after a GPL statement thereby leading to the conclusion that it is their INTENTION to GPL the firmware.
If we have a source:
--
/* This file is under the GPL, yada yada */ #include "things.h"
void some_func(void) { does_something(); }
char firmware[]={0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07};
void upload_firmware(void) { do_upload(firmware); }
--
Then it seems clear to me that the firmware is under the GPL because it is PART of the GPL'd file. If not, then I don't see how any statement can ever be true to similar effect, even for some_func().
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