Xref: utzoo comp.lang.c:6252 comp.sys.m68k:654
Path: utzoo!utgpu!water!watmath!clyde!rutgers!mit-eddie!apollo!johnf
From: johnf@apollo.uucp (John Francis)
Newsgroups: comp.lang.c,comp.sys.m68k
Subject: Re: C machine
Message-ID: <39aca826.7f32@apollo.uucp>
Date: 14 Jan 88 19:42:00 GMT
References: <461@auvax.UUCP> <9961@mimsy.UUCP> <166@teletron.UUCP> <6936@brl-smoke.ARPA> <147@ateng.UUCP>
Organization: Apollo Computer, Chelmsford, Mass.
Lines: 25


>Why would a compiler vendor for the 68000 choose a 32 bit size for an int?

Consider:

    main()
        {
        char    c[100000],p1,p2;
        int     dp;

        p1 = &c[0];
        p2 = &c[99999];

        dp = p2 - p1;
        }

The result of subtracting two pointers is *defined* (K&R) to yield
a result of type int. Not long int - int!  Any implementation that
can not represent this in an int is not an implementation of C.

Mind you, any decent 68000 compiler should provide a 16-bit short int,
and the code generator should be able to handle a = b + c (all shorts)
without converting b & c to ints, adding them, and then converting the
result to short.  If it is really good it should be able to do the same
for a = b + 2.