static void
BBox_Cubic_Check( FT_Pos y1,
FT_Pos y2,
FT_Pos y3,
FT_Pos y4,
FT_Pos* min,
FT_Pos* max )
{
/* always compare first and last points */
if ( y1 < *min ) *min = y1;
else if ( y1 > *max ) *max = y1;
if ( y4 < *min ) *min = y4;
else if ( y4 > *max ) *max = y4;
/* now, try to see if there are split points here */
if ( y1 <= y4 )
{
/* flat or ascending arc test */
if ( y1 <= y2 && y2 <= y4 && y1 <= y3 && y3 <= y4 )
return;
}
else /* y1 > y4 */
{
/* descending arc test */
if ( y1 >= y2 && y2 >= y4 && y1 >= y3 && y3 >= y4 )
return;
}
/* There are some split points. Find them. */
{
FT_Pos a = y4 - 3*y3 + 3*y2 - y1;
FT_Pos b = y3 - 2*y2 + y1;
FT_Pos c = y2 - y1;
FT_Pos d;
FT_Fixed t;
/* We need to solve `ax^2+2bx+c' here, without floating points! */
/* The trick is to normalize to a different representation in order */
/* to use our 16.16 fixed point routines. */
/* */
/* We compute FT_MulFix(b,b) and FT_MulFix(a,c) after normalization. */
/* These values must fit into a single 16.16 value. */
/* */
/* We normalize a, b, and c to `8.16' fixed float values to ensure */
/* that its product is held in a `16.16' value. */
{
FT_ULong t1, t2;
int shift = 0;
/* The following computation is based on the fact that for */
/* any value `y', if `n' is the position of the most */
/* significant bit of `abs(y)' (starting from 0 for the */
/* least significant bit), then `y' is in the range */
/* */
/* -2^n..2^n-1 */
/* */
/* We want to shift `a', `b', and `c' concurrently in order */
/* to ensure that they all fit in 8.16 values, which maps */
/* to the integer range `-2^23..2^23-1'. */
/* */
/* Necessarily, we need to shift `a', `b', and `c' so that */
/* the most significant bit of its absolute values is at */
/* _most_ at position 23. */
/* */
/* We begin by computing `t1' as the bitwise `OR' of the */
/* absolute values of `a', `b', `c'. */
t1 = (FT_ULong)( ( a >= 0 ) ? a : -a );
t2 = (FT_ULong)( ( b >= 0 ) ? b : -b );
t1 |= t2;
t2 = (FT_ULong)( ( c >= 0 ) ? c : -c );
t1 |= t2;
/* Now we can be sure that the most significant bit of `t1' */
/* is the most significant bit of either `a', `b', or `c', */
/* depending on the greatest integer range of the particular */
/* variable. */
/* */
/* Next, we compute the `shift', by shifting `t1' as many */
/* times as necessary to move its MSB to position 23. This */
/* corresponds to a value of `t1' that is in the range */
/* 0x40_0000..0x7F_FFFF. */
/* */
/* Finally, we shift `a', `b', and `c' by the same amount. */
/* This ensures that all values are now in the range */
/* -2^23..2^23, i.e., they are now expressed as 8.16 */
/* fixed-float numbers. This also means that we are using */
/* 24 bits of precision to compute the zeros, independently */
/* of the range of the original polynomial coefficients. */
/* */
/* This algorithm should ensure reasonably accurate values */
/* for the zeros. Note that they are only expressed with */
/* 16 bits when computing the extrema (the zeros need to */
/* be in 0..1 exclusive to be considered part of the arc). */
if ( t1 == 0 ) /* all coefficients are 0! */
return;
if ( t1 > 0x7FFFFFUL )
{
do
{
shift++;
t1 >>= 1;
} while ( t1 > 0x7FFFFFUL );
/* this loses some bits of precision, but we use 24 of them */
/* for the computation anyway */
a >>= shift;
b >>= shift;
c >>= shift;
}
else if ( t1 < 0x400000UL )
{
do
{
shift++;
t1 <<= 1;
} while ( t1 < 0x400000UL );
a <<= shift;
b <<= shift;
c <<= shift;
}
}
/* handle a == 0 */
if ( a == 0 )
{
if ( b != 0 )
{
t = - FT_DivFix( c, b ) / 2;
test_cubic_extrema( y1, y2, y3, y4, t, min, max );
}
}
else
{
/* solve the equation now */
d = FT_MulFix( b, b ) - FT_MulFix( a, c );
if ( d < 0 )
return;
if ( d == 0 )
{
/* there is a single split point at -b/a */
t = - FT_DivFix( b, a );
test_cubic_extrema( y1, y2, y3, y4, t, min, max );
}
else
{
/* there are two solutions; we need to filter them */
d = FT_SqrtFixed( (FT_Int32)d );
t = - FT_DivFix( b - d, a );
test_cubic_extrema( y1, y2, y3, y4, t, min, max );
t = - FT_DivFix( b + d, a );
test_cubic_extrema( y1, y2, y3, y4, t, min, max );
}
}
}
static void
BBox_Cubic_Check( FT_Pos y1,
FT_Pos y2,
FT_Pos y3,
FT_Pos y4,
FT_Pos* min,
FT_Pos* max )
{
/* always compare first and last points */
if ( y1 < *min ) *min = y1;
else if ( y1 > *max ) *max = y1;
if ( y4 < *min ) *min = y4;
else if ( y4 > *max ) *max = y4;
/* now, try to see if there are split points here */
if ( y1 <= y4 )
{
/* flat or ascending arc test */
if ( y1 <= y2 && y2 <= y4 && y1 <= y3 && y3 <= y4 )
return;
}
else /* y1 > y4 */
{
/* descending arc test */
if ( y1 >= y2 && y2 >= y4 && y1 >= y3 && y3 >= y4 )
return;
}
/* There are some split points. Find them. */
{
FT_Pos a = y4 - 3*y3 + 3*y2 - y1;
FT_Pos b = y3 - 2*y2 + y1;
FT_Pos c = y2 - y1;
FT_Pos d;
FT_Fixed t;
/* We need to solve "ax^2+2bx+c" here, without floating points! */
/* The trick is to normalize to a different representation in order */
/* to use our 16.16 fixed point routines. */
/* */
/* We compute FT_MulFix(b,b) and FT_MulFix(a,c) after the */
/* the normalization. These values must fit into a single 16.16 */
/* value. */
/* */
/* We normalize a, b, and c to "8.16" fixed float values to ensure */
/* that their product is held in a "16.16" value. */
/* */
{
FT_ULong t1, t2;
int shift = 0;
/* Technical explanation of what's happening there. */
/* */
/* The following computation is based on the fact that for */
/* any value "y", if "n" is the position of the most */
/* significant bit of "abs(y)" (starting from 0 for the */
/* least significant bit), then y is in the range */
/* */
/* "-2^n..2^n-1" */
/* */
/* We want to shift "a", "b" and "c" concurrently in order */
/* to ensure that they all fit in 8.16 values, which maps */
/* to the integer range "-2^23..2^23-1". */
/* */
/* Necessarily, we need to shift "a", "b" and "c" so that */
/* the most significant bit of their absolute values is at */
/* _most_ at position 23. */
/* */
/* We begin by computing "t1" as the bitwise "or" of the */
/* absolute values of "a", "b", "c". */
/* */
t1 = (FT_ULong)((a >= 0) ? a : -a );
t2 = (FT_ULong)((b >= 0) ? b : -b );
t1 |= t2;
t2 = (FT_ULong)((c >= 0) ? c : -c );
t1 |= t2;
/* Now, the most significant bit of "t1" is sure to be the */
/* msb of one of "a", "b", "c", depending on which one is */
/* expressed in the greatest integer range. */
/* */
/* We now compute the "shift", by shifting "t1" as many */
/* times as necessary to move its msb to position 23. */
/* */
/* This corresponds to a value of t1 that is in the range */
/* 0x40_0000..0x7F_FFFF. */
/* */
/* Finally, we shift "a", "b" and "c" by the same amount. */
/* This ensures that all values are now in the range */
/* -2^23..2^23, i.e. that they are now expressed as 8.16 */
/* fixed float numbers. */
/* */
/* This also means that we are using 24 bits of precision */
/* to compute the zeros, independently of the range of */
/* the original polynom coefficients. */
/* */
/* This should ensure reasonably accurate values for the */
/* zeros. Note that the latter are only expressed with */
/* 16 bits when computing the extrema (the zeros need to */
/* be in 0..1 exclusive to be considered part of the arc). */
/* */
if ( t1 == 0 ) /* all coefficients are 0! */
return;
if ( t1 > 0x7FFFFFUL )
{
do
{
shift++;
t1 >>= 1;
} while ( t1 > 0x7FFFFFUL );
/* losing some bits of precision, but we use 24 of them */
/* for the computation anyway. */
a >>= shift;
b >>= shift;
c >>= shift;
}
else if ( t1 < 0x400000UL )
{
do
{
shift++;
t1 <<= 1;
} while ( t1 < 0x400000UL );
a <<= shift;
b <<= shift;
c <<= shift;
}
}
/* handle a == 0 */
if ( a == 0 )
{
if ( b != 0 )
{
t = - FT_DivFix( c, b ) / 2;
test_cubic_extrema( y1, y2, y3, y4, t, min, max );
}
}
else
{
/* solve the equation now */
d = FT_MulFix( b, b ) - FT_MulFix( a, c );
if ( d < 0 )
return;
if ( d == 0 )
{
/* there is a single split point at -b/a */
t = - FT_DivFix( b, a );
test_cubic_extrema( y1, y2, y3, y4, t, min, max );
}
else
{
/* there are two solutions; we need to filter them though */
d = FT_SqrtFixed( (FT_Int32)d );
t = - FT_DivFix( b - d, a );
test_cubic_extrema( y1, y2, y3, y4, t, min, max );
t = - FT_DivFix( b + d, a );
test_cubic_extrema( y1, y2, y3, y4, t, min, max );
}
}
}
