Technique: Bit count & bitset

Bit Count

	#include <bits/stdc++.h>
	using namespace std;
	int main() {
	// 4 bit
	for (int i = 0; i < (1 << 4); i++) {
	// convert to 8 bit
	cout << i << "(" << bitset<8>(i) << ")";
	// count bits
	cout << ": " << __builtin_popcount(i) << " bits." << endl;
	}
	return 0;
	}
	/*
	0(00000000): 0 bits.
	1(00000001): 1 bits.
	2(00000010): 1 bits.
	3(00000011): 2 bits.
	4(00000100): 1 bits.
	5(00000101): 2 bits.
	6(00000110): 2 bits.
	7(00000111): 3 bits.
	8(00001000): 1 bits.
	9(00001001): 2 bits.
	10(00001010): 2 bits.
	11(00001011): 3 bits.
	12(00001100): 2 bits.
	13(00001101): 3 bits.
	14(00001110): 3 bits.
	15(00001111): 4 bits.
	*/

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bitset

	#include <iostream>
	#include <bitset>
	int main()
	{
	// 整数から8ビットのビット集合を構築
	std::bitset<8> bs1(131uL); // 10000011
	// 文字列から8ビットのビット集合を構築
	std::bitset<8> bs2("10000011");
	// 1ビット目が1かを判定
	if (bs1[1]) {
	std::cout << "1st bit is 1" << std::endl;
	}
	// 2ビット目を1にする
	bs1.set(2);
	std::cout << "2nd bit to 1 : " << bs1 << std::endl;
	// 2ビット目を0に戻す
	bs1.reset(2);
	// いずれかのビットが1かを判定
	if (bs1.any()) {
	std::cout << "some bits are 1" << std::endl;
	}
	// 論理演算
	std::bitset<8> and_bits = bs1 & std::bitset<8>("10000001"); // 論理積
	std::bitset<8> or_bits = bs1 | std::bitset<8>("00010100"); // 論理和
	std::bitset<8> xor_bits = bs1 ^ std::bitset<8>("00100011"); // 排他的論理和
	std::cout << "and : " << and_bits << std::endl;
	std::cout << "or : " << or_bits << std::endl;
	std::cout << "xor : " << xor_bits << std::endl;
	}
	/*
	1st bit is 1
	2nd bit to 1 : 10000111
	some bits are 1
	and : 10000001
	or : 10010111
	xor : 10100000
	*/

view raw Bit Set.cpp hosted with ❤ by GitHub