1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
|
ARM QEMU doesn't enforce that RES0 bits in FPSCR are non-writeable
Hi all, we systematically tested the QEMU implementation for emulating arm user mode programs. We found that QEMU incorrectly emulate the FPSCR register. The following the proof of code:
/*********** Beginning of the bug: arm.c **********/
int printf(const char *format, ...);
unsigned char i0[0x10];
unsigned char o[0x10];
int main() {
int k = 0;
asm("mov r2, %0\n"
"ldr r0, [r2]\n"::"r"((char *)(i0)));;
asm("vmsr fpscr, r0");
asm("mov r2, %0\n"
"vmrs r4, fpscr\n"
"str r4, [r2]\n"::"r"((char *)(o)));;
for (k = 0; k < 0x10; k++)
printf("%02x", o[0x10 - 1 - k]);
printf("\n");
}
unsigned char i0[0x10] = {0xff, 0xff, 0xff, 0xff, 0x00, 0x00, 0x00, 0x00, 0x28, 0x1c, 0xc7, 0x01, 0x00, 0x00, 0x00, 0x00};
/*********** End fo the bug **********/
When the program is compiled into arm binary code and running on a real arm machine, and running in qemu, we have the following result
$ arm-linux-gnueabihf-gcc arm.c -o arm -static
$ ./arm
000000000000000000000000fff7009f
$ qemu-arm arm
000000000000000000000000ffffffff
According to the ARM manual, bits[19, 14:13, 6:5] of FPSCR should be reserved as zero. However, arm qemu fails to keep these bits to be zero: these bits can be actually modified in QEMU.
Thanks!
|