In the CMSIS definitions for gcc you can find something like this:
static __INLINE void __DMB(void) { __ASM volatile ("dmb"); }
My question is: what use does a memory barrier have if it does not declare "memory" in the clobber list?
Is it an error in the core_cm3.h or is there a reason why gcc should behave correctly without any additional help?
I did some testing with gcc 4.5.2 (built with LTO). If I compile this code:
static inline void __DMB(void) { asm volatile ("dmb"); }
static inline void __DMB2(void) { asm volatile ("dmb" ::: "memory"); }
char x;
char test1 (void)
{
x = 15;
return x;
}
char test2 (void)
{
x = 15;
__DMB();
return x;
}
char test3 (void)
{
x = 15;
__DMB2();
return x;
}
using arm-none-eabi-gcc -Os -mcpu=cortex-m3 -mthumb -c dmb.c
, then from arm-none-eabi-objdump -d dmb.o
I get this:
00000000 <test1>:
0: 4b01 ldr r3, [pc, #4] ; (8 <test1+0x8>)
2: 200f movs r0, #15
4: 7018 strb r0, [r3, #0]
6: 4770 bx lr
8: 00000000 .word 0x00000000
0000000c <test2>:
c: 4b02 ldr r3, [pc, #8] ; (18 <test2+0xc>)
e: 200f movs r0, #15
10: 7018 strb r0, [r3, #0]
12: f3bf 8f5f dmb sy
16: 4770 bx lr
18: 00000000 .word 0x00000000
0000001c <test3>:
1c: 4b03 ldr r3, [pc, #12] ; (2c <test3+0x10>)
1e: 220f movs r2, #15
20: 701a strb r2, [r3, #0]
22: f3bf 8f5f dmb sy
26: 7818 ldrb r0, [r3, #0]
28: 4770 bx lr
2a: bf00 nop
2c: 00000000 .word 0x00000000
It is obvious that __DBM()
only inserts the dmb
instruction and it takes DMB2()
to actually force the compiler to flush the values cached in the registers.
I guess I found a CMSIS bug.
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