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msm1089 avatar msm1089 commented on May 27, 2024

UPDATE

In the end I solved it by changing the last else block:

    else:
        start_mask = (1 << (8 - start_bit)) - 1
        end_mask = ~((1 << (8 - end_bit - 1)) - 1)
        screen[start_byte] |= start_mask
        screen[end_byte] = int('11111111', base=2)
        screen[end_byte] &= end_mask

I looked back at the first challenge (bits) in bit_manipulation series. I set the last bit to all 1s, then used the method from the clear_bits_index_to_lsb method in the bits challenge.

I have a couple of questions about this.

  1. Setting all bits in a byte is done using int('11111111', base=2). This could just be done by setting it to equal 255. What is the reason to use the much longer way?
  2. Is there some mask that can be used to set all bits from MSB to an index? This would prevent the need to first set all bits then clear the bits to the right of the index.

from interactive-coding-challenges.

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