Question

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?

Answer 1

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.