On 4/27/05, Ken Hirsch <[log in to unmask]> wrote:

>This is not true of C in general, just C on most machines.  The C
>standard requires those casts for a reason. From an old USENET C FAQ:

OK - I stand corrected.  I've never used any of those other machines.

<snip>
>The old HP 3000 series uses a different addressing scheme for byte
>addresses than for word addresses; like several of the machines above it
>therefore uses different representations for char * and void * pointers
>than for other pointers.

That may be the case but it doesn't manifest itself (easily) when you use
the C compiler on that system.  I'd be interested to see a program that
demonstrates it.  This is my attempt -


#include <stdio.h>
#include <memory.h>

main(int argc, char *argv[])
{
        int *mybuff = (int *)malloc(100);
        /* Assign from int * to char * without and with casting */
        char *c_mybuff1 = mybuff;
        char *c_mybuff2 = (char *)mybuff;

        /* Put something in mybuff */
        strcpy((char *)mybuff, "Hello!");

        /* The following displays Hello! twice. */
        printf("c_mybuff1 = %s\n", c_mybuff1);
        printf("c_mybuff2 = %s\n", c_mybuff2);
}

$ make Z9
c89 -O  -o Z9 Z9.c
cc: "/DEV/DEV/PETERS/Z9.c", line 8: warning 604: Pointers are not
assignment-com
patible.

$ ./Z9
c_mybuff1 = Hello!
c_mybuff2 = Hello!
$ uname -a
MPE/iX QAHP3K  C.70.00 C.39.06  SERIES 979-200
$

Maybe I need to use an older version of MPE or another compiler?  I tried
using "short" instead of "int" so that it was a nice 16 bit word.  Trouble
is, I'm compiling 32 bit so this must complicate matters.  Maybe I need to
compile in 16 bit mode?  I'm not sure how to do that.

Cheers.

Peter
--
http://www.beluga.freeserve.co.uk

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