Messages in this thread | | | Date | Thu, 17 Jan 2008 15:10:13 -0800 (PST) | From | Linus Torvalds <> | Subject | RE: Why is the kfree() argument const? |
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On Thu, 17 Jan 2008, David Schwartz wrote: > > No, that's not what it means. It has nothing to do with memory. It has to do > with logical state.
Blah. That's just your own made-up explanation of what you think "const" should mean. It has no logical background or any basis in the C language.
"const" has nothing to do with "logical state". It has one meaning, and one meaning only: the compiler should complain if that particular type is used to do a write access.
It says nothing at all about the "logical state of the object". It cannot, since a single object can - and does - have multiple pointers to it.
So your standpoint not only has no relevant background to it, it's also not even logically consistent.
Linus
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