我有一个C ++中的自定义位集类实现。我经常遍历位集中设置的位。该迭代可以如下实现:
struct Bitset {
uint64_t* data_;
size_t chunks_;
std::vector<int> Elements() const {
std::vector<int> ret;
for (size_t i=0;i<chunks_;i++){
uint64_t td = data_[i];
while (td) {
ret.push_back(i*BITS + __builtin_ctzll(td));
td &= ~-td;
}
}
return ret;
}
};
void Iterate(Bitset bitset) {
for (int b : bitset.Elements()) {
std::cout << "bit: " << b << std::endl;
}
}
上面的实现为迭代提供了干净的代码,但是它涉及到向量不必要的堆分配。基本上内联Elements()函数的以下版本通常更快:
void Iterate(Bitset bitset) {
int chunks = bitset.chunks_;
for (int i = 0; i < chunks; i++) {
uint64_t td = bitset.data_[i];
while (td) {
std::cout << "bit: " << i*BITS + __builtin_ctzll(td) << std::endl;
td &= ~-td;
}
}
}
为迭代实现抽象的一种好方法,使它与上述版本一样干净,而且没有性能损失。
只需遍历您的课程。为Bitset
提供您自己的迭代器类的实现,并提供begin()
和end()
方法。最简单(未经测试!)的实现可能如下所示:
#include <vector>
#include <cstdint>
#include <iostream>
struct Bitset {
uint64_t* data_;
size_t chunks_;
struct iterator {
uint64_t *pnt;
uint_fast8_t pos;
iterator(uint64_t *pnt, size_t pos) :
pnt(pnt), pos(pos) {}
bool operator !=(const iterator& o) {
return o.pnt != pnt || o.pos != pos;
}
void operator ++() {
pos++;
if (pos == 64) {
pnt++;
pos = 0;
}
}
bool operator *() {
return *pnt & (1 << pos);
}
};
iterator begin() { return iterator(data_, 0); }
iterator end() { return iterator(data_ + chunks_, 64); }
};
void Iterate(Bitset bitset) {
for (auto&& b : bitset) {
std::cout << "bit: " << b << std::endl;
}
}
我相信您不奇怪的while (td) { ... i*BITS + __builtin_ctzll(td) ...
循环,可能会一直存在(未经测试!):
constexpr int BITS = 100000;
struct Bitset {
uint64_t* data_;
size_t chunks_;
struct iterator {
uint64_t *data_;
int i = 0;
uint64_t td = 0;
iterator(uint64_t *data_, int i, uint64_t td) :
data_(data_), i(i), td(td) {}
bool operator !=(const iterator& o) {
return o.data_ != data_ || o.i != i || o.td != td;
}
void operator ++() {
if (td == 0) {
td = *data_;
data_++;
} else {
td &= ~-td;
}
}
bool operator *() {
return i * BITS + __builtin_ctzll(td);
}
};
iterator begin() { return iterator(data_, 0, *data_); }
iterator end() { return iterator(data_ + chunks_, 0, 0); }
};