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From: mbrennan@swift.cs.tcd.ie
Newsgroups: comp.sys.apollo
Subject: Question on MMAP(2)
Message-ID: <1990Nov28.172733.7570@swift.cs.tcd.ie>
Date: 28 Nov 90 17:27:32 GMT
Organization: Computer Science Department, Trinity College Dublin
Lines: 35

Can anyone explain to me how the mmap(2) system call actually works.
I know that it maps an area from a file into virtual memory, but I
am not sure of some of the finer details of the relationship between 
the size of the file as it appears in the file system and the size of 
the mmap'ed area?

For example I have a file which is 300 bytes long (created using vi).
When I do an mmap call I must specify 4096 bytes (on the DN10000) as
the length of the segment to be mapped (len must be a multiple of the
pagesize).  Sure enough the return value in len indicates that 300 bytes 
have been mapped.

However if I copy say a 1000 byte string (using strcpy) into that mapped
region there is no objection.  If I terminate the program and examine the 
data file then I can see the first 300 bytes of the string I copied into 
the mapped region - fair enough.  If I rerun the program, map 4096 bytes 
from the same 300-byte data file I can now access the 1000 byte string I 
copied in there during the last run of the program.  My question is 
where were bytes 301-1000 of that string stored between successive runs
of the program - or maybe more correctly what was I clobbering.

The following is my mmap call:
	len = 4096 ;
	ret = mmap(NULL, &len, PROT_READ | PROT_WRITE, 
		MAP_FILE | MAP_NOEXTEND | MAP_SHARED, fd, 0) ;

It is being run on an
	Apollo DN10000 under DomainOS 10.2 (bsd 4.3)   and/or
	Apollo 3550 under DomainOS 10.1 (bsd 4.3)

Any help at all appreciated.
--
 ,   ,  ,      ,              ,       ,          ,          , ,      ,
Micheal O Braonain     Roinn Riomheolaiochta, Colaiste Na Trinoide, BAC 2.
Email                  mbrennan@cs.tcd.ie