Write bits in different bytes c c++ -

this question exact duplicate of:

• extract , combine bits different bytes c c++ 1 answer

i have declared array of bytes:

uint8_t memory[123]; 

which have filled with:

memory[0]=0xff; memory[1]=0x00; memory[2]=0xff; memory[3]=0x00; memory[4]=0xff; 

and requests user write in bits. example, user provides starting bit (ex:10), amount of bits (ex:9) , bits set. in example provided receive 2 bytes:

setbit[0]=0b11110010; setbit[1]=0b00000001; //padded zeros bits 

this used modbus big-endian protocol. have come following code:

for(int j=findbyteinit;j<(findbytefinal);j++){     aux[0]=(unsigned char) (setbit[j]>>(startingbit-(8*findbyteinit)));    aux[1]=(unsigned char) (setbit[j+1]<<(startingbit-(8*findbyteinit)));     memory[j]=(unsigned char) (aux[0] & memory[j] );    memory[j+1]=(unsigned char) (aux[1] & memory[j+1] );      aux[0]=0x00;//clean aux    aux[1]=0x00;          } 

which not work should close ideal solution. suggestions?

i think should work. i've tested lightly.

  int start_bit = 10; int bit_count = 9;   uint8_t setbit[2] = { 0b11110010, 0b00000001 };    int setbit_size = (bit_count + (char_bit - 1)) / char_bit;    int start_byte = start_bit / char_bit;   int shift = start_bit % char_bit;   int modified_bytes =       bit_count ? (bit_count + shift + (char_bit - 1)) / char_bit : 0;    (int = 0; < modified_bytes; ++i) {     uint8_t mask = 0xff;     if (i == 0) {       mask <<= shift;     } else if (i == modified_bytes - 1) {       mask >>= char_bit - (bit_count + shift) % char_bit;     }      uint8_t carried = > 0 ? (setbit[i - 1] >> (char_bit - shift)) : 0;     uint8_t added = < setbit_size ? static_cast<uint8_t>(setbit[i] << shift) : 0;     memory[start_byte + i] =         (memory[start_byte + i] & ~mask) | ((carried | added) & mask);   }