旋转等待原子变成一定值

时间:2019-04-02 18:15:48

标签: c assembly stdatomic

我想旋转一个原子,使其变为某个值,然后获取与商店存储memory_order_release在一起的内存。

我认为通过优化

#include <stdatomic.h>

void wait_for_flag_to_become2(_Atomic unsigned char *xp)
{
    while(2!=atomic_load_explicit(xp,memory_order_acquire));
}
void wait_for_flag_to_become2_(_Atomic unsigned char *xp)
{
    while(2!=atomic_load_explicit(xp,memory_order_relaxed)){}
    atomic_thread_fence(memory_order_acquire);
}

将生成相同的代码,并且该代码将类似于后者的功能,但是在输出程序集不同的平台上,它们不会。

以上两个函数在语义上是否相等,并且是由于简单的优化失败而导致输出程序集的差异,还是实际上在语义上不相等?


https://gcc.godbolt.org/z/hNcf-i

power64-at12,gcc8:

wait_for_flag_to_become2:
        .quad   .L.wait_for_flag_to_become2,.TOC.@tocbase,0
.L.wait_for_flag_to_become2:
.L2:
        lbz 9,0(3)
        cmpw 7,9,9
        bne- 7,$+4
        isync
        rlwinm 9,9,0,0xff
        cmplwi 7,9,2
        bne 7,.L2
        blr
        .long 0
        .byte 0,0,0,0,0,0,0,0
wait_for_flag_to_become2_:
        .quad   .L.wait_for_flag_to_become2_,.TOC.@tocbase,0
.L.wait_for_flag_to_become2_:
.L6:
        lbz 9,0(3)
        cmplwi 7,9,2
        bne 7,.L6
        lwsync
        blr
        .long 0
        .byte 0,0,0,0,0,0,0,0

ARM64,gcc 8.2:

wait_for_flag_to_become2:
.L2:
        ldarb   w1, [x0]
        and     w1, w1, 255
        cmp     w1, 2
        bne     .L2
        ret
wait_for_flag_to_become2_:
.L5:
        ldrb    w1, [x0]
        and     w1, w1, 255
        cmp     w1, 2
        bne     .L5
        dmb     ishld
        ret

0 个答案:

没有答案