On Tue, Nov 20, 2001 at 11:54:13PM +0900, Bang Jun-Young wrote:

 > On Tue, Nov 20, 2001 at 03:32:02AM -0500, der Mouse wrote:
 > > > 	mmap((void *)0x50000000, 4096, PROT_READ, 0, fd, 0);
 > > > 	mmap((void *)0x50000000, 4096, PROT_READ, MAP_FIXED, fd, 0);
 > > > 	mmap((void *)0x50000000, 4096, PROT_READ, 0, fd, 0);
 > > > 	mmap((void *)0x50000000, 4096, PROT_READ, MAP_FIXED, fd, 0);
 > > 
 > > > Here I expected the second and the fourth mmap()'s would fail, since
 > > > there was a mapping already done at 0x50000000 by the first mmap.
 > > > Even if the second succeeded, I expected at least the fourth would
 > > > fail.  The result was that they _all_ succeded.
 > > 
 > > mmap(2) - at least on my machines - says, in part,
 > > 
 > > A successful
 > > .Fa mmap
 > > deletes any previous mapping in the allocated address range.
 > > 
 > > This seems to me to be a fairly clear statement that if you mmap
 > > something on top of something else, the previous mapping is, to the
 > > extent that it conflicts with the new one, unmapped and replaced.
 > 
 > At first I expected that, but the result shows that mmap(2) doesn't
 > act like that: the third mmap call returned 0x50001000, not 0x50000000.
 > If the manpage was right, mmap should zap the previous mapping at
 > 0x50000000.

No, because you did not use MAP_FIXED.  The first argument to mmap is
only used as a hint if MAP_FIXED is not supplied...

-- 
        -- Jason R. Thorpe <thorpej@wasabisystems.com>