oral_quiz->#整数二进制后，1的数目#

#include <stdio.h>

//太渣， 负数还不行
int MyNumberOf1InBinary(int num) {
	int numOfZero = 0;
	int divided = num;
	while(divided != 0) {
		if(divided % 2 == 1) ++numOfZero;	// ==1 都不需要
		divided /= 2;
	}

	return numOfZero;
}

//negative drop-dead halt
int NumberOf1_s1(int n) {
	int count = 0;
	while(n) {
		if(n & 1) {
			++count;
		}
		n = n >> 1;
	}
	return count;
}

//solution2
int NumberOf1_s2(int n) {
	int count = 0;
	unsigned int flag = 1;
	while(flag) {
		if(n & flag) {
			++count;
		}
		flag = flag << 1;
	}
	return count;
}

//good solution3
int NumberOf1_s3(int n) {
	int count = 0;
	while(n) {
		++count;
		n &= n - 1;
	}
	return count;
}

void Test(int num, int expected) {
	printf("Test for %d ", num);
	if(expected == NumberOf1_s3(num))
		printf("passed.\n");
	else
		printf("failed.\n");
}

int main(int argc, char* argv[]) {
	Test(9, 2);
	Test(7 ,3);
	Test(4, 1);
	Test(0, 0);
	Test(-9, 31);
	Test(0x7FFFFFFF, 31);
	Test(0x80000000, 1);
	printf("%d\n", 0x80000000 - 1);
	return 0;
}