在 stackoverflow 上有這么[1]一個(gè)問題，問題的答案中有這么幾段：

At the same time, x86 defines quite a strict memory model, which bans most possible reorderings, roughly summarized as follows:

Stores have a single global order of visibility, observed consistently by all CPUs, subject to one loosening of this rule below. Local load operations are never reordered with respect to other local load operations.

Local store operations are never reordered with respect to other local store operations (i.e., a store that appears earlier in the instruction stream always appears earlier in the global order).

Local load operations may be reordered with respect to earlier local store operations, such that the load appears to execute earlier wrt the global store order than the local store, but the reverse (earlier load, older store) is not true.

簡(jiǎn)單概括一下，就是在 x86 平臺(tái)采用較強(qiáng)的內(nèi)存序，只有 store load 會(huì)發(fā)生亂序。

看各位八股文老仙們背的實(shí)在辛苦，本文提供一點(diǎn)可以直接實(shí)操證明這些問題的手段。

perfbook 一書在講 memory barrier 相關(guān)的概念時(shí)，都使用了一個(gè)叫 litmus 的工具，現(xiàn)在被集成在?herdtools[2]?中，安裝好 herdtools 就已經(jīng)有了 litmus，上面提到的所有讀寫重排/亂序的情況我們都可以進(jìn)行測(cè)試。

讀寫亂序測(cè)試

X86?RW {?x=0;?y=0;?}P0??????????|?P1??????????;MOV?EAX,[y]?|?MOV?EAX,[x]?;MOV?[x],$1??|?MOV?[y],$1??; locations?[x;y;] exists?(0:EAX=1?/\?1:EAX=1) %%%%%%%%%%%%%%%%%%%%%%%%% %?Results?for?sb.litmus?% %%%%%%%%%%%%%%%%%%%%%%%%% X86?OOO{x=0;?y=0;}P0??????????|?P1??????????;MOV?EAX,[y]?|?MOV?EAX,[x]?;MOV?[x],$1??|?MOV?[y],$1??;locations?[x;?y;] exists?(0:EAX=1?/\?1:EAX=1) Generated?assembler##START?_litmus_P0movl?-4(%rsi,%rcx,4),?%eaxmovl?$1,?-4(%rbx,%rcx,4)##START?_litmus_P1movl?-4(%rbx,%rcx,4),?%eaxmovl?$1,?-4(%rsi,%rcx,4)Test?OOO?Allowed Histogram?(2?states) 500000:>0:EAX=1;?1:EAX=0;?x=1;?y=1; 500000:>0:EAX=0;?1:EAX=1;?x=1;?y=1; NoWitnesses Positive:?0,?Negative:?1000000 Condition?exists?(0:EAX=1?/\?1:EAX=1)?is?NOT?validated Hash=7cdd62e8647b817c1615cf8eb9d2117b Observation?OOO?Never?0?1000000 Time?OOO?0.14

寫讀亂序測(cè)試

X86?RW {?x=0;?y=0;?}P0??????????|?P1??????????;MOV?EAX,[y]?|?MOV?EAX,[x]?;MOV?[x],$1??|?MOV?[y],$1??; locations?[x;y;] exists?(0:EAX=1?/\?1:EAX=1) %%%%%%%%%%%%%%%%%%%%%%%%%% %?Results?for?sb2.litmus?% %%%%%%%%%%%%%%%%%%%%%%%%%% X86?OOO{x=0;?y=0;}P0??????????|?P1??????????;MOV?[x],$1??|?MOV?[y],$1??;MOV?EAX,[y]?|?MOV?EAX,[x]?;locations?[x;?y;] exists?(0:EAX=0?/\?1:EAX=0) Generated?assembler##START?_litmus_P0movl?$1,?-4(%rbx,%rcx,4)movl?-4(%rsi,%rcx,4),?%eax##START?_litmus_P1movl?$1,?-4(%rsi,%rcx,4)movl?-4(%rbx,%rcx,4),?%eaxTest?OOO?Allowed Histogram?(4?states) 2?????*>0:EAX=0;?1:EAX=0;?x=1;?y=1; 499998:>0:EAX=1;?1:EAX=0;?x=1;?y=1; 499999:>0:EAX=0;?1:EAX=1;?x=1;?y=1; 1?????:>0:EAX=1;?1:EAX=1;?x=1;?y=1; OkWitnesses Positive:?2,?Negative:?999998 Condition?exists?(0:EAX=0?/\?1:EAX=0)?is?validated Hash=2d53e83cd627ba17ab11c875525e078b Observation?OOO?Sometimes?2?999998 Time?OOO?0.12

讀讀和寫寫亂序測(cè)試

這里我沒想到太好的辦法，所以將讀讀和寫寫混在一起進(jìn)行測(cè)試，無論是 WW 會(huì)發(fā)生重排，或是 RR 會(huì)發(fā)生重排，都可能會(huì)出現(xiàn)在 P0 中，EAX = 2，EBX = 0 的情況。

X86?OOO {?x=0;?y=0;?}P0??????????|?P1??????????;MOV?EAX,[x]?|?MOV?[y],$1??;MOV?EBX,[y]?|?MOV?[x],$2??; locations?[x;y;] exists?(0:EAX=2?/\?0:EBX=0) %%%%%%%%%%%%%%%%%%%%%%%%%% %?Results?for?sb3.litmus?% %%%%%%%%%%%%%%%%%%%%%%%%%% X86?OOO{x=0;?y=0;}P0??????????|?P1?????????;MOV?EAX,[x]?|?MOV?[y],$1?;MOV?EBX,[y]?|?MOV?[x],$2?;locations?[x;?y;] exists?(0:EAX=2?/\?0:EBX=0) Generated?assembler##START?_litmus_P0movl?-4(%rbx,%rcx,4),?%eaxmovl?-4(%rdx,%rcx,4),?%r11d##START?_litmus_P1movl?$1,?-4(%rdi,%rax,4)movl?$2,?-4(%rcx,%rax,4)Test?OOO?Allowed Histogram?(3?states) 500000:>0:EAX=0;?0:EBX=0;?x=2;?y=1; 1?????:>0:EAX=0;?0:EBX=1;?x=2;?y=1; 499999:>0:EAX=2;?0:EBX=1;?x=2;?y=1; NoWitnesses Positive:?0,?Negative:?1000000 Condition?exists?(0:EAX=2?/\?0:EBX=0)?is?NOT?validated Hash=74f6930f2a61d6cfec9fb5ea3132555e Observation?OOO?Never?0?1000000 Time?OOO?0.11

[1]

這么:?https://stackoverflow.com/questions/50307693/does-an-x86-cpu-reorder-instructions

[2]

herdtools:?https://github.com/herd/herdtools7

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