kernel oops due to unaligned access with lswi
Olaf Hering
olh at suse.de
Sun Nov 16 08:04:49 EST 2003
Alan,
2.6 has a function parse_header(), its part of gunzip and used for
initramfs as example. I got an oops, exception 600, sig 7. I think that
means unaligned access.
Why does the compiler optimize this one? I see it with 2.95, 3.2, 3.3
and 3.4 with -O1 and better, not with -O0.
gcc -msoft-float -mmultiple -mstring -O2 -Wall -Wstrict-prototypes \
-Wno-trigraphs -Wno-uninitialized -version -fno-strict-aliasing \
-fno-common -ffixed-r2 -fomit-frame-pointer -o parse_header \
-c parse_header.c -v --save-temps
lswi 9,31,8
stswi 9,28,8
s = r31. How can gcc be sure that s aligned?
ppc64 will call memcpy (tested with gcc3.2.3).
.file "parse_header.c"
.section .init.text,"ax", at progbits
.align 2
.type parse_header, at function
parse_header:
stwu 1,-112(1)
mflr 0
stmw 27,92(1)
stw 0,116(1)
li 0,0
stb 0,64(1)
li 30,0
addi 31,3,6
addi 28,1,56
addi 27,1,8
.L6:
lswi 9,31,8
stswi 9,28,8
slwi 29,30,2
mr 3,28
li 4,0
li 5,16
bl simple_strtoul
stwx 3,29,27
addi 30,30,1
addi 31,31,8
cmpwi 0,30,11
ble+ 0,.L6
lwz 0,116(1)
mtlr 0
lmw 27,92(1)
addi 1,1,112
blr
.Lfe1:
.size parse_header,.Lfe1-parse_header
.section ".text"
.align 2
.globl main
.type main, at function
main:
stwu 1,-16(1)
mflr 0
stw 0,20(1)
lis 3,0x1
ori 3,3,57920
bl malloc
bl parse_header
li 3,0
lwz 0,20(1)
mtlr 0
addi 1,1,16
blr
.Lfe2:
.size main,.Lfe2-main
.ident "GCC: (GNU) 3.2.3 (SuSE Linux)"
#include <stdlib.h>
typedef unsigned int __kernel_size_t;
extern void *memcpy(void *, const void *, __kernel_size_t);
extern unsigned long simple_strtoul(const char *, char **, unsigned int);
static void __attribute__ ((__section__(".init.text"))) parse_header(char *s)
{
unsigned long parsed[12];
char buf[9];
int i;
buf[8] = '\0';
for (i = 0, s += 6; i < 12; i++, s += 8) {
memcpy(buf, s, 8);
parsed[i] = simple_strtoul(buf, ((void *) 0), 16);
}
}
int
main(void)
{
char *s;
s = malloc(123456);
parse_header(s);
return 0;
}
--
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