以下程序中位掩码的使用来自《Programming Pearls》
我今天开始阅读“编程珍珠”，在做练习时我遇到了这个问题“你将如何实现你自己的位向量？”。当我查看解决方案时，它是这样的：
#define BITSPERWORD 32
#define SHIFT 5
#define MASK 0x1F
#define N 10000000
int a[1 + N/BITSPERWORD];
void set(int i) { a[i >> SHIFT] |= (1 << (i & MASK));
我对这个声明感到困惑
1 << (i & MASK)
有人可以解释一下这里发生了什么吗？
I started reading "Programming Pearls" today and while doing it's exercise I came across this question "How would you implement your own bit vector?". When I looked at the solution it was like this:
#define BITSPERWORD 32
#define SHIFT 5
#define MASK 0x1F
#define N 10000000
int a[1 + N/BITSPERWORD];
void set(int i) { a[i >> SHIFT] |= (1 << (i & MASK));
Where I am getting confused at is this statement
1 << (i & MASK)
Could someone please explain me what's going on here?
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请注意，
MASK的设置使其具有最低SHIFT位设置，其中SHIFT恰好是BITSPERWORD 的以 2 为底的对数。因此
(i & MASK)会选择i的最低5位，相当于除以32后取余（只要考虑如何取最低两位即可）例如，十进制数的位数给出除以 100 后的余数）。这给出了我们感兴趣的单词内的位数。1 << (i & MASK))（顺便说一下，这是一个表达式，而不是语句）现在创建一个值，其中恰好是我们'重新感兴趣已设置。使用|=将此值合并到内存字中将设置位向量的所需位。Note that
MASKis set such that it has the lowestSHIFTbits set, whereSHIFTis exactly the base-2 logarithm ofBITSPERWORD.Therefore
(i & MASK)will select the lowest 5 bits ofi, which is the same as taking the remainder after dividing by 32 (just consider how taking the lowest two digits of a decimal number gives you the remainder after dividing by 100, for example). That gives the number of the bit within a word we're interested in.1 << (i & MASK))(which is, by the way, an expression, not a statement) now creates a value where exactly the bit we're interested in is set. Merging this value into the memory word with|=will set the desired bit of the bit vector.0x20 是 32，所以 i & 0x1F 采用
i模 32，因此您永远不会移位 32 位。这是一种保护措施，因为移动任何不严格小于类型大小的内容都是未定义的行为。0x20 is 32, so
i & 0x1Ftakesimodulo 32, so that you never shift by 32 bits. This is a safeguard because shifting by anything that isn't strictly less than the size of the type is undefined behaviour.