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Full Discussion: 0x20 and such
Top Forums Programming 0x20 and such Post 302123467 by porter on Monday 25th of June 2007 08:55:47 PM
Old 06-25-2007
Quote:
Originally Posted by Octal
Or maybe not? I know that the bit 7 in there is the 'case bit', which can turn on/off weather the character is capital. So I'm assuming you can use other bits in this fashion?
Not quite,

In Ascii, the lower case are 'a' to 'z', which are 0x61 to 0x7A, and upper case are 0x41 to 0x5A, so you could treat bit 6 as a case bit if the character falls in this range, but this falls down when you add diacriticals.

There are the the << and >> operators, which will shift a value left or right a number of bits. Hence (x)<<7 would convert 0x01 to 0x40 for example.

Note that the least significant bit is consider "bit zero", and has a value of 1. If you set "bit one", you would have a value of two.
 
BITMAP_ONTO(9)						  Basic Kernel Library Functions					    BITMAP_ONTO(9)

NAME
bitmap_onto - translate one bitmap relative to another SYNOPSIS
void bitmap_onto(unsigned long * dst, const unsigned long * orig, const unsigned long * relmap, int bits); ARGUMENTS
dst resulting translated bitmap orig original untranslated bitmap relmap bitmap relative to which translated bits number of bits in each of these bitmaps DESCRIPTION
Set the n-th bit of dst iff there exists some m such that the n-th bit of relmap is set, the m-th bit of orig is set, and the n-th bit of relmap is also the m-th _set_ bit of relmap. (If you understood the previous sentence the first time your read it, you're overqualified for your current job.) In other words, orig is mapped onto (surjectively) dst, using the the map { <n, m> | the n-th bit of relmap is the m-th set bit of relmap }. Any set bits in orig above bit number W, where W is the weight of (number of set bits in) relmap are mapped nowhere. In particular, if for all bits m set in orig, m >= W, then dst will end up empty. In situations where the possibility of such an empty result is not desired, one way to avoid it is to use the bitmap_fold operator, below, to first fold the orig bitmap over itself so that all its set bits x are in the range 0 <= x < W. The bitmap_fold operator does this by setting the bit (m % W) in dst, for each bit (m) set in orig. Example [1] for bitmap_onto: Let's say relmap has bits 30-39 set, and orig has bits 1, 3, 5, 7, 9 and 11 set. Then on return from this routine, dst will have bits 31, 33, 35, 37 and 39 set. When bit 0 is set in orig, it means turn on the bit in dst corresponding to whatever is the first bit (if any) that is turned on in relmap. Since bit 0 was off in the above example, we leave off that bit (bit 30) in dst. When bit 1 is set in orig (as in the above example), it means turn on the bit in dst corresponding to whatever is the second bit that is turned on in relmap. The second bit in relmap that was turned on in the above example was bit 31, so we turned on bit 31 in dst. Similarly, we turned on bits 33, 35, 37 and 39 in dst, because they were the 4th, 6th, 8th and 10th set bits set in relmap, and the 4th, 6th, 8th and 10th bits of orig (i.e. bits 3, 5, 7 and 9) were also set. When bit 11 is set in orig, it means turn on the bit in dst corresponding to whatever is the twelfth bit that is turned on in relmap. In the above example, there were only ten bits turned on in relmap (30..39), so that bit 11 was set in orig had no affect on dst. Example [2] for bitmap_fold + bitmap_onto: Let's say relmap has these ten bits set: 40 41 42 43 45 48 53 61 74 95 (for the curious, that's 40 plus the first ten terms of the Fibonacci sequence.) Further lets say we use the following code, invoking bitmap_fold then bitmap_onto, as suggested above to avoid the possitility of an empty dst result: unsigned long *tmp; // a temporary bitmap's bits bitmap_fold(tmp, orig, bitmap_weight(relmap, bits), bits); bitmap_onto(dst, tmp, relmap, bits); Then this table shows what various values of dst would be, for various orig's. I list the zero-based positions of each set bit. The tmp column shows the intermediate result, as computed by using bitmap_fold to fold the orig bitmap modulo ten (the weight of relmap). orig tmp dst 0 0 40 1 1 41 9 9 95 10 0 40 (*) 1 3 5 7 1 3 5 7 41 43 48 61 0 1 2 3 4 0 1 2 3 4 40 41 42 43 45 0 9 18 27 0 9 8 7 40 61 74 95 0 10 20 30 0 40 0 11 22 33 0 1 2 3 40 41 42 43 0 12 24 36 0 2 4 6 40 42 45 53 78 102 211 1 2 8 41 42 74 (*) (*) For these marked lines, if we hadn't first done bitmap_fold into tmp, then the dst result would have been empty. If either of orig or relmap is empty (no set bits), then dst will be returned empty. If (as explained above) the only set bits in orig are in positions m where m >= W, (where W is the weight of relmap) then dst will once again be returned empty. All bits in dst not set by the above rule are cleared. COPYRIGHT
Kernel Hackers Manual 3.10 June 2014 BITMAP_ONTO(9)
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