計算十進制數中二進制1的個數

方便查找，來源見參考文檔

你學的C是什麼C，你說的++又是什麼++

1

while(sum>0)  {  
    if(sum%2 != 0)  {  
        c++;   // counting number of ones  
    }  
    sum=sum/2;  
}  

2

#include <bitset>
#include <iostream>
#include <climits>

size_t popcount(size_t n) {
    std::bitset<sizeof(size_t) * CHAR_BIT> b(n);
    return b.count();
}

int main() {
    std::cout << popcount(1000000);
}

3

#ifdef __APPLE__
#define NAME(name) _##name
#else
#define NAME(name) name
#endif

/*
 * Count the number of bits set in the bitboard.
 *
 * %rdi: bb
 */
.globl NAME(cpuPopcount);
NAME(cpuPopcount):
    popcnt %rdi, %rax
    ret
/*
 * Test if the CPU has the popcnt instruction.
 */
.globl NAME(cpuHasPopcount);
NAME(cpuHasPopcount):
    pushq %rbx

    movl $1, %eax
    cpuid                   // ecx=feature info 1, edx=feature info 2

    xorl %eax, %eax

    testl $1 << 23, %ecx
    jz 1f
    movl $1, %eax

1:
    popq %rbx
    ret
unsigned cppPopcount(unsigned bb)
{
#define C55 0x5555555555555555ULL
#define C33 0x3333333333333333ULL
#define C0F 0x0f0f0f0f0f0f0f0fULL
#define C01 0x0101010101010101ULL

    bb -= (bb >> 1) & C55;              // put count of each 2 bits into those 2 bits
    bb = (bb & C33) + ((bb >> 2) & C33);// put count of each 4 bits into those 4 bits
    bb = (bb + (bb >> 4)) & C0F;        // put count of each 8 bits into those 8 bits
    return (bb * C01) >> 56;            // returns left 8 bits of x + (x<<8) + (x<<16) + (x<<24) + ...
}
unsigned builtinPopcount(unsigned bb)
{
    return __builtin_popcountll(bb);
}

3

unsigned int v = value(); // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
for (c = 0; v; c++)
{
  v &= v - 1; // clear the least significant bit set
}

4

 int number = 15; // this is input number
    int oneCount = number & 1 ? 1 : 0;
    while(number = number >> 1)
    {
        if(number & 1)
            ++oneCount;
    }
    cout << "# of ones :"<< oneCount << endl;

5

int count_1s_in_Num(int num)
{
    int count=0;
    while(num!=0)
    {
        num = num & (num-1);
        count++;
    }
    return count;
}

“If you apply the AND operation to the integer and the result of the subtraction, the result is a new number that is the same as the original integer except that the rightmost 1 is now a 0. For example,01110000 AND (01110000 – 1) = 01110000 AND 01101111 = 01100000.This solution has a running time of O(m), where m is the number of 1s in the solution.”
“如果你對減法的結果和一個整數使用與操作，結果會是和原始整數一樣的一個新數，除了最右邊的1此時爲0。結果時間複雜度爲O(m)，m爲數中1的個數。”

參考文檔

[1].https://stackoverflow.com/questions/14682641/count-number-of-1s-in-binary-format-of-decimal-number/14682688#14682688