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Home/ Questions/Q 6987629
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T18:57:44+00:00 2026-05-27T18:57:44+00:00

i have a three dimensional array which i am using as a bit table

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i have a three dimensional array which i am using as a bit table

   char bit_table_[X][Y][Z];

X is not larger than 20 but Y and Z will be very large.
the contents of every X’s Y and Z will be compared parallely as follows (here real values of Y and Z will be computed using some hash functions).
My problem is; I don’t know if there is any way at all to tell as to which of the X’s give true in the condition checking of the if statment

if (((bit_table_[0][i][bit_index / bits_per_char]|
bit_table_[1][i][bit_index / bits_per_char])& bit_mask[bit]) != bit_mask[bit])
     return true;

or is there anyother way of doing it?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T18:57:45+00:00Added an answer on May 27, 2026 at 6:57 pm

    You have to test them individually to know which one resulted in true. Here is an example using a logical rather than bitwise OR to determine which one resulted in true.

    bool x1 = false, x2 = false;
    
    if(
       (x1 = (bit_table_[0][i][bit_index/bits_per_char] & bit_mask[bit]) != bit_mask[bit]) ||
       (x2 = (bit_table_[1][i][bit_index/bits_per_char] & bit_mask[bit]) != bit_mask[bit])
      )
    {
        //Check x1 and x2 here
        return true;
    }
    

    Edit:

    To expand on the previous example, and to satisfy the results of your original post you could also check if the combination of both is the reason it passes like so:

    bool x1 = false, x2 = false, both = false;
    size_t zindex = bit_index/bits_per_char;
    if(
       (x1 = (bit_table_[0][i][zindex] & bit_mask[bit]) != bit_mask[bit]) ||
       (x2 = (bit_table_[1][i][zindex] & bit_mask[bit]) != bit_mask[bit]) ||
       (both = ((bit_table_[0][i][zindex] | bit_table_[1][i][zindex]) &
                bit_mask[bit]) != bit_mask[bit])
      )
    {
        //Check x1 and x2 here
        //If both is true then neither x1 or x2 resulted in true alone
    
        return true;
    }
    
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