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Home/ Questions/Q 7612727
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T01:59:10+00:00 2026-05-31T01:59:10+00:00

I have the following: #define PAD ( 4 – ( (WIDTH*BPP)%4 ) ) #if

  • 0

I have the following:

#define PAD (  4 - ( (WIDTH*BPP)%4 )  )
#if PAD == 4
#define PAD 0
#endif

and PAD is redefined even though it is equal to 3 after the first definition. However if I explicitly define it as 3 then it isn’t redefined. Therefore I assume there is a problem with the way I have written the expression, but I’m not sure what.

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  1. Editorial Team
    Editorial Team
    2026-05-31T01:59:12+00:00Added an answer on May 31, 2026 at 1:59 am

    What you want is

    (PAD + (WIDTH * BPP)) % 4 == 0
    

    right? (Of course 0 <= PAD < 4)

    Then you can define PAD in this way:

    #define PAD (3 - ((WIDTH * BPP + 3) % 4))
    

    Example Python session:

    >>> def f(x): return 3 - (x+3)%4
    ...
    >>> [ (x, f(x), x + f(x)) for x in xrange(100,108) ]
    [(100, 0, 100), (101, 3, 104), (102, 2, 104), (103, 1, 104), (104, 0, 104), (105, 3, 108), (106, 2, 108), (107, 1, 108)]
    

    In general,

    #define PAD ((N-1) - (X + (N-1)) % N))
    

    makes PAD + X a multiple of N under a constraint of 0 <= PAD < N (Though I didn’t check negative cases…)

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