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Although this appears to be harmless, it is possible (and likely) that flag1 and flag2 are stored in the same byte. If both assignments occur on a thread scheduling interleaving that ends with both stores occurring after one another, it is possible that only one of the flags will be set as intended and the other flag will equal its previous value. This is because both bit-fields are represented by the same byte, which is the smallest unit the processor could can work on.
For example, the following sequence of events could can occur.
| Code Block |
|---|
Thread 1: register 0 = flags Thread 1: register 0 &= ~mask(flag1) Thread 2: register 0 = flags Thread 2: register 0 &= ~mask(flag2) Thread 1: register 0 |= 1 << shift(flag1) Thread 1: flags = register 0 Thread 2: register 0 |= 2 << shift(flag2) Thread 2: flags = register 0 |
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