在STM32上指定位置沿数组的有效位计算

Question

I'm wondering if someone know effective approach to calculate bits in specified position along array?

Answer 1

You can just loop the array values and test for the bits with a bitwise and operator, like so:

int arr[] = {1,2,3,4,5};
// 1 - 001
// 2 - 010
// 3 - 011
// 4 - 100
// 5 - 101

int i, bitcount = 0;

for (i = 0; i < 5; ++i){
    if (arr[i] & (1 << 2)){ //testing and counting the 3rd bit
        bitcount++;
    }
}

printf("%d", bitcount); //2

Note that i opted for 1 << 2 which tests for the 3rd bit from the right or the third least significant bit just to be easier to show. Now bitCount would now hold 2 which are the number of 3rd bits set to 1 .

Take a look at the result in Ideone

In your case you would need to check for the 5th bit which can be represented as:

• 1 << 4
• 0x10000
• 16

And the 8th bit:

• 1 << 7
• 0x10000000
• 256

So adjusting this to your bits would give you:

int i, bitcount8 = 0, bitcount5 = 0;

for (i = 0; i < your_array_size_here; ++i){
    if (arr[i] & 0x10000000){
        bitcount8++;
    }
    if (arr[i] & 0x10000){
        bitcount5++;
    }
}

If you need to count many of them, then this solution isn't great and you'd be better off creating an array of bit counts, and calculating them with another for loop:

int i, j, bitcounts[8] = {0};

for (i = 0; i < your_array_size_here; ++i){
    for (j = 0; j < 8; ++j){
        //j will be catching each bit with the increasing shift lefts
        if (arr[i] & (1 << j)){
            bitcounts[j]++;
        }
    }
}

And in this case you would access the bit counts by their index:

printf("%d", bitcounts[2]); //2

Check this solution in Ideone as well

Answer 2

Assuming that OP wants to count active bits

size_t countbits(uint8_t *array, int pos, size_t size)
{
    uint8_t mask = 1 << pos;
    uint32_t result = 0;

    while(size--)
    {
        result += *array++ & mask;
    }
    return result >> pos;
}

Answer 3

Let the bit position difference (eg 7 - 4 in this case) be diff .

If 2 ^diff > n, then code can add both bits at the same time.

void count(const uint8_t *Array, size_t n, int *bit7sum, int *bit4sum) {
  unsigned sum = 0;
  unsigned mask = 0x90;
  while (n > 0) {
    n--;
    sum += Array[n] & mask;
  }
  *bit7sum = sum >> 7;
  *bit4sum = (sum >> 4) & 0x07;
}

---

If the processor has a fast multiply and n is still not too large, like n < pow(2,14) in this case. (Or n < pow(2,8) in the general case)

void count2(const uint8_t *Array, size_t n, int *bit7sum, int *bit4sum) {
  // assume 32 bit or wider unsigned
  unsigned sum = 0;
  unsigned mask1 = 0x90;
  unsigned m = 1 + (1u << 11);  // to move bit 7 to the bit 18 place
  unsigned mask2 = (1u << 18) | (1u << 4);
  while (n > 0) {
    n--;
    sum += ((Array[n] & mask1)*m) & mask2;
  }
  *bit7sum = sum >> 18;
  *bit4sum = ((1u << 18) - 1) & sum) >> 4);
}

Algorithm: code is using a mask, multiply, mask to separate the 2 bits. The lower bit remains in it low position while the upper bit is shifted to the upper bits. Then a parallel add occurs.

The loop avoids any branching aside from the loop itself. This can make for fast code. YMMV .

With even larger n , break it down into multiple calls to count2()