In these two macro, what is the "per_cpu__##var"? TIA
#define per_cpu(var, cpu) (*((void)cpu, &per_cpu__##var)) #define __get_cpu_var(var) per_cpu__##var
Hi, ## is a concatenation operator. So, IF you pass, __get_cpu_var(x), it becomes,
per_cpu_x
It is used for get percpu things.
regards manish
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