Linux cpumask分析

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首先我们来分析一下定义在cpumask.h中结构体cpumask_t

typedef struct cpumask { DECLARE_BITMAP(bits, NR_CPUS); } cpumask_t;#define DECLARE_BITMAP(name,bits) \unsigned long name[BITS_TO_LONGS(bits)]#define BITS_TO_LONGS(nr)DIV_ROUND_UP(nr, BITS_PER_BYTE * sizeof(long))#define DIV_ROUND_UP(n,d) (((n) + (d) - 1) / (d))

假设我们当前使用的cpu核数为24，即NR_CPUS=24，sizeof(long)=8，BITS_PER_BYTE=8，则在DIV_ROUND_UP(n,d)中， n = 24,d=64，宏的展开结果为(24+64-1)/64 = 1，DECLARE_BITMAP(name,bits) 展开后即为 unsigned long name[1]，最后cpumask即为：

struct cpumask{

unsigned long bits[1];

};

绕了好大一个圈，就定义了一个unsigned long bits[1]，想想就明白了，64位机器上，一个long有64个bit，而只有24个核，所以一个long足够表示了。在linux内核中，cpu_possible_mask 位图，用来表示系统中的CPU，每颗处理器对应其中一位，
cpu_online_mask 位图，用来当前处于工作状态的CPU，每颗处理器对应其中一位

接下来，分析cpu_bit_map

/* * cpu_bit_bitmap[] is a special, "compressed" data structure that * represents all NR_CPUS bits binary values of 1<<nr. * * It is used by cpumask_of() to get a constant address to a CPU * mask value that has a single bit set only. *//* cpu_bit_bitmap[0] is empty - so we can back into it */#define MASK_DECLARE_1(x)[x+1][0] = (1UL << (x))#define MASK_DECLARE_2(x)MASK_DECLARE_1(x), MASK_DECLARE_1(x+1)#define MASK_DECLARE_4(x)MASK_DECLARE_2(x), MASK_DECLARE_2(x+2)#define MASK_DECLARE_8(x)MASK_DECLARE_4(x), MASK_DECLARE_4(x+4)const unsigned long cpu_bit_bitmap[BITS_PER_LONG+1][BITS_TO_LONGS(NR_CPUS)] = {MASK_DECLARE_8(0),MASK_DECLARE_8(8),MASK_DECLARE_8(16),MASK_DECLARE_8(24),#if BITS_PER_LONG > 32MASK_DECLARE_8(32),MASK_DECLARE_8(40),MASK_DECLARE_8(48),MASK_DECLARE_8(56),#endif};EXPORT_SYMBOL_GPL(cpu_bit_bitmap);

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