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@ -357,7 +357,7 @@ |
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FT_Fixed t; |
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/* we need to solve "ax^2+2bx+c" here, without floating points! */ |
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/* We need to solve "ax^2+2bx+c" here, without floating points! */ |
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/* The trick is to normalize to a different representation in order */ |
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/* to use our 16.16 fixed point routines. */ |
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/* */ |
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@ -373,25 +373,25 @@ |
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int shift = 0; |
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/* technical explanation of what's happening there */ |
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/* Technical explanation of what's happening there. */ |
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/* */ |
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/* the following computation is based on the fact that for */ |
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/* The following computation is based on the fact that for */ |
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/* any value "y", if "n" is the position of the most */ |
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/* significant bit of "abs(y)" (starting from 0 for the */ |
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/* least significant bit), then y is in the range */ |
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/* */ |
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/* "-2^n..2^n-1" */ |
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/* */ |
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/* we want to shift "a", "b" and "c" concurrently in order */ |
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/* We want to shift "a", "b" and "c" concurrently in order */ |
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/* to ensure that they all fit in 8.16 values, which maps */ |
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/* to the integer range "-2^23..2^23-1" */ |
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/* to the integer range "-2^23..2^23-1". */ |
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/* */ |
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/* necessarily, we need to shift "a", "b" and "c" so that */ |
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/* Necessarily, we need to shift "a", "b" and "c" so that */ |
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/* the most significant bit of their absolute values is at */ |
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/* _most_ at position 23 */ |
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/* _most_ at position 23. */ |
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/* */ |
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/* we begin by computing "t1" as the bitwise "or" of the */ |
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/* absolute values of "a", "b", "c" */ |
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/* We begin by computing "t1" as the bitwise "or" of the */ |
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/* absolute values of "a", "b", "c". */ |
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/* */ |
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t1 = (FT_ULong)((a >= 0) ? a : -a ); |
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t2 = (FT_ULong)((b >= 0) ? b : -b ); |
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@ -399,31 +399,30 @@ |
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t2 = (FT_ULong)((c >= 0) ? c : -c ); |
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t1 |= t2; |
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/* now, the most significant bit of "t1" is sure to be the */ |
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/* Now, the most significant bit of "t1" is sure to be the */ |
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/* msb of one of "a", "b", "c", depending on which one is */ |
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/* expressed in the greatest integer range.. */ |
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/* expressed in the greatest integer range. */ |
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/* */ |
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/* we will now compute the "shift", by shifting "t1" as many */ |
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/* We now compute the "shift", by shifting "t1" as many */ |
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/* times as necessary to move its msb to position 23. */ |
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/* */ |
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/* this corresponds to a value of t1 that is in the range */ |
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/* 0x40_0000..0x7F_FFFF */ |
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/* This corresponds to a value of t1 that is in the range */ |
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/* 0x40_0000..0x7F_FFFF. */ |
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/* */ |
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/* finally, we shift "a", "b" and "c" by the same amount. */ |
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/* this ensure that all values are now in the range */ |
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/* -2^23..2^23, i.e. that they're now expressed as 8.16 */ |
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/* fixed float numbers.. */ |
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/* Finally, we shift "a", "b" and "c" by the same amount. */ |
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/* This ensures that all values are now in the range */ |
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/* -2^23..2^23, i.e. that they are now expressed as 8.16 */ |
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/* fixed float numbers. */ |
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/* */ |
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/* this also means that we're using 24 bits of precision */ |
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/* This also means that we are using 24 bits of precision */ |
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/* to compute the zeros, independently of the range of */ |
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/* the original polynom coefficients. */ |
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/* */ |
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/* this should ensure reasonably accurate values for the */ |
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/* zeros. Note that the latter are only expressed with */ |
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/* This should ensure reasonably accurate values for the */ |
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/* zeros. Note that the latter are only expressed with */ |
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/* 16 bits when computing the extrema (the zeros need to */ |
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/* be in 0..1 exclusive to be considered part of the arc) */ |
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/* be in 0..1 exclusive to be considered part of the arc). */ |
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/* */ |
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if ( t1 == 0 ) /* all coefficients are 0! */ |
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return; |
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@ -435,8 +434,8 @@ |
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t1 >>= 1; |
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} while ( t1 > 0x7FFFFFUL ); |
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/* losing some bits of precision, but we'll use 24 of them */ |
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/* for the computation anyway.. */ |
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/* losing some bits of precision, but we use 24 of them */ |
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/* for the computation anyway. */ |
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a >>= shift; |
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b >>= shift; |
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c >>= shift; |
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Werner Lemberg
24 years ago