> Based on that, neither 0 nor 1 is defined. Yes, it is. But I think the comment is to describe the return value of numa_initialize() not for numa_available(). That is also the reason why it has 0 and 1 as its return values. Oscar
> Based on that, neither 0 nor 1 is defined. Yes, it is. But I think the comment is to describe the return value of numa_initialize() not for numa_available(). That is also the reason why it has 0 and 1 as its return values. Oscar
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