Addresses with 4GB differences are consider as one single address in QEMU

[Khansa Butt]

THIS IS THE ISSUE OF USER MODE EMULATION
Information about guest and host
**********************************
guest: 64 bit x86 user mode binary
host: 32 bit Linux OS
uname -a :Linux KICS-HPCNL-32blue 2.6.33.3-85.fc13.i686.PAE #1 SMP
architecture: intel64
Bug Description
****************
for memory reference instructions, suppose I have two addresses in guest address space(64 bit)
0x220000000
0x320000000
as lower 32 bit part of both addresses are same, when particular instructions are translated into host code(32 bit)
in both above cases the value is loaded from same memory and we get same value. where actual behaviour was to get two different values.
here is the program which i used to test:
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#define SIZE 4294967297 /* 4Gib*/

int main() {
   char *array;
   unsigned int i;

   array = malloc(sizeof(char) * SIZE);
   if(array == NULL) {
      fprintf(stderr, "Could not allocate that much memory");
      return 1; }
    array[0] = 'a';
   array[SIZE-1] = 'z';
   printf("array[SIZE-1] = %c array[0] = %c\n",array[SIZE-1], array[0]);
  return 0;
}
I have 8 gib RAM
I compiled this program on 64 bit linux and run this on 32 bit linux with qemu
QEMU command line and output
**********************************
$x86_64-linux-user/qemu-x86_64 ~/ar_x86
output: array[SIZE-1] = z,array[0] = z
Release information
********************
x86_64 binary is tested with latest release : qemu-0.14.1
and with current development tree as well( live code of QEMU using git)

[Thomas Huth (th-huth)]

Can you still reproduce this problem with the latest version of QEMU (currently version 2.9.0)?

[Launchpad Janitor (janitor)]

[Expired for QEMU because there has been no activity for 60 days.]