位操作，保存位，8位后扩展

Question

I want to traverse a binary tree from the bottom to the top.

Then i want to save the bitsequence of this traversion(=the way) in a char.

This behaviour should be dynamic, so if i have a bitsequence of more than 8 bits, the char should be dynamically expanded, for example, 2 bytes and so on....

If the bitsequence is for example 1001010 i want that exact same bitsequence stored inside the char.

I know that i should use the bitshift operators << >> but i can't quite figure out the correct way to do it.

After i wrote 8 bits in the char i encounter a problem.

I attached some sample code, hopefully somebody can shed some light.

Thanks

char* bits = malloc(sizeof(char));
char* temp_bits = NULL;

some loop
{
  if (cnt_bit > 7)
  {
    temp_bits = realloc(bits, sizeof(char)*2);
    free(bits);
    bits = temp_bits;
  }
  *bits = *bits << 1;
  *bits = *bits | 0;
  cnt_bit++;
}

Answer 1

You can't shift from one memory value to the next! If you want to do that you could try to use the type "long long int" which is 64 bit normally but i don't think you can go further than that with shifts, unless you implement your own shift operation which works on a generic array.

I think what you are doing would be much faster by using one byte per level.