On Mon, 13 Jun 2011, Khansa Butt wrote:
> Public bug reported:
>
> 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 4294967298 /* 4Gib*/
>
> int main() {
Invalid declaration of main.
> char *array;
> unsigned int i;
i is never used.
>
> array = malloc(sizeof(char) * SIZE);
sizeof(char) is by definition 1.
> if(array == NULL) {
> fprintf(stderr, "Could not allocate that much memory");
This lacks line termination.
> return 1; }
> array[0] = 'a';
> array[SIZE-2] = 'z';
> printf("array[SIZE-2] = %c array[0] = %c\n",array[SIZE-2], 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
And this doesn't match what you printf at all (it's spelled "SIZE-2" in
the format string in the code snippet above), but since main is misdeclared
(hello UB) all bets are off and nasal monsters are to be expected anyways.
> 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)
>
> ** Affects: qemu
> Importance: Undecided
> Status: New
>
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