/*
* tan(x) = sin(x) / cos(x)
*/
struct fpn *
fpu_tan(struct fpemu *fe)
{
struct fpn x;
struct fpn s;
struct fpn *r;
if (ISNAN(&fe->fe_f2))
return &fe->fe_f2;
if (ISINF(&fe->fe_f2))
return fpu_newnan(fe);
/* if x is +0/-0, return +0/-0 */
if (ISZERO(&fe->fe_f2))
return &fe->fe_f2;
CPYFPN(&x, &fe->fe_f2);
/* sin(x) */
CPYFPN(&fe->fe_f2, &x);
r = fpu_sin(fe);
CPYFPN(&s, r);
/* cos(x) */
CPYFPN(&fe->fe_f2, &x);
r = fpu_cos(fe);
CPYFPN(&fe->fe_f2, r);
CPYFPN(&fe->fe_f1, &s);
r = fpu_div(fe);
return r;
}
/*
* arccos(x) = pi/2 - arcsin(x)
*/
struct fpn *
fpu_acos(struct fpemu *fe)
{
struct fpn *r;
if (ISNAN(&fe->fe_f2))
return &fe->fe_f2;
if (ISINF(&fe->fe_f2))
return fpu_newnan(fe);
r = fpu_asin(fe);
CPYFPN(&fe->fe_f2, r);
/* pi/2 - asin(x) */
fpu_const(&fe->fe_f1, FPU_CONST_PI);
fe->fe_f1.fp_exp--;
fe->fe_f2.fp_sign = !fe->fe_f2.fp_sign;
r = fpu_add(fe);
return r;
}
/*
* x
* arcsin(x) = arctan(---------------)
* sqrt(1 - x^2)
*/
struct fpn *
fpu_asin(struct fpemu *fe)
{
struct fpn x;
struct fpn *r;
if (ISNAN(&fe->fe_f2))
return &fe->fe_f2;
if (ISZERO(&fe->fe_f2))
return &fe->fe_f2;
if (ISINF(&fe->fe_f2))
return fpu_newnan(fe);
CPYFPN(&x, &fe->fe_f2);
/* x^2 */
CPYFPN(&fe->fe_f1, &fe->fe_f2);
r = fpu_mul(fe);
/* 1 - x^2 */
CPYFPN(&fe->fe_f2, r);
fe->fe_f2.fp_sign = 1;
fpu_const(&fe->fe_f1, FPU_CONST_1);
r = fpu_add(fe);
/* sqrt(1-x^2) */
CPYFPN(&fe->fe_f2, r);
r = fpu_sqrt(fe);
/* x/sqrt */
CPYFPN(&fe->fe_f2, r);
CPYFPN(&fe->fe_f1, &x);
r = fpu_div(fe);
/* arctan */
CPYFPN(&fe->fe_f2, r);
return fpu_atan(fe);
}
/*
* Our task is to calculate the square root of a floating point number x0.
* This number x normally has the form:
*
* exp
* x = mant * 2 (where 1 <= mant < 2 and exp is an integer)
*
* This can be left as it stands, or the mantissa can be doubled and the
* exponent decremented:
*
* exp-1
* x = (2 * mant) * 2 (where 2 <= 2 * mant < 4)
*
* If the exponent `exp' is even, the square root of the number is best
* handled using the first form, and is by definition equal to:
*
* exp/2
* sqrt(x) = sqrt(mant) * 2
*
* If exp is odd, on the other hand, it is convenient to use the second
* form, giving:
*
* (exp-1)/2
* sqrt(x) = sqrt(2 * mant) * 2
*
* In the first case, we have
*
* 1 <= mant < 2
*
* and therefore
*
* sqrt(1) <= sqrt(mant) < sqrt(2)
*
* while in the second case we have
*
* 2 <= 2*mant < 4
*
* and therefore
*
* sqrt(2) <= sqrt(2*mant) < sqrt(4)
*
* so that in any case, we are sure that
*
* sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2
*
* or
*
* 1 <= sqrt(n * mant) < 2, n = 1 or 2.
*
* This root is therefore a properly formed mantissa for a floating
* point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2
* as above. This leaves us with the problem of finding the square root
* of a fixed-point number in the range [1..4).
*
* Though it may not be instantly obvious, the following square root
* algorithm works for any integer x of an even number of bits, provided
* that no overflows occur:
*
* let q = 0
* for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
* x *= 2 -- multiply by radix, for next digit
* if x >= 2q + 2^k then -- if adding 2^k does not
* x -= 2q + 2^k -- exceed the correct root,
* q += 2^k -- add 2^k and adjust x
* fi
* done
* sqrt = q / 2^(NBITS/2) -- (and any remainder is in x)
*
* If NBITS is odd (so that k is initially even), we can just add another
* zero bit at the top of x. Doing so means that q is not going to acquire
* a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the
* final value in x is not needed, or can be off by a factor of 2, this is
* equivalant to moving the `x *= 2' step to the bottom of the loop:
*
* for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
*
* and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
* (Since the algorithm is destructive on x, we will call x's initial
* value, for which q is some power of two times its square root, x0.)
*
* If we insert a loop invariant y = 2q, we can then rewrite this using
* C notation as:
*
* q = y = 0; x = x0;
* for (k = NBITS; --k >= 0;) {
* #if (NBITS is even)
* x *= 2;
* #endif
* t = y + (1 << k);
* if (x >= t) {
* x -= t;
* q += 1 << k;
* y += 1 << (k + 1);
* }
* #if (NBITS is odd)
* x *= 2;
* #endif
* }
*
* If x0 is fixed point, rather than an integer, we can simply alter the
* scale factor between q and sqrt(x0). As it happens, we can easily arrange
* for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
*
* In our case, however, x0 (and therefore x, y, q, and t) are multiword
* integers, which adds some complication. But note that q is built one
* bit at a time, from the top down, and is not used itself in the loop
* (we use 2q as held in y instead). This means we can build our answer
* in an integer, one word at a time, which saves a bit of work. Also,
* since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
* `new' bits in y and we can set them with an `or' operation rather than
* a full-blown multiword add.
*
* We are almost done, except for one snag. We must prove that none of our
* intermediate calculations can overflow. We know that x0 is in [1..4)
* and therefore the square root in q will be in [1..2), but what about x,
* y, and t?
*
* We know that y = 2q at the beginning of each loop. (The relation only
* fails temporarily while y and q are being updated.) Since q < 2, y < 4.
* The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
* Furthermore, we can prove with a bit of work that x never exceeds y by
* more than 2, so that even after doubling, 0 <= x < 8. (This is left as
* an exercise to the reader, mostly because I have become tired of working
* on this comment.)
*
* If our floating point mantissas (which are of the form 1.frac) occupy
* B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
* In fact, we want even one more bit (for a carry, to avoid compares), or
* three extra. There is a comment in fpu_emu.h reminding maintainers of
* this, so we have some justification in assuming it.
*/
struct fpn *
fpu_sqrt(struct fpemu *fe)
{
struct fpn *x = &fe->fe_f1;
u_int bit, q, tt;
u_int x0, x1, x2, x3;
u_int y0, y1, y2, y3;
u_int d0, d1, d2, d3;
int e;
FPU_DECL_CARRY;
/*
* Take care of special cases first. In order:
*
* sqrt(NaN) = NaN
* sqrt(+0) = +0
* sqrt(-0) = -0
* sqrt(x < 0) = NaN (including sqrt(-Inf))
* sqrt(+Inf) = +Inf
*
* Then all that remains are numbers with mantissas in [1..2).
*/
DPRINTF(FPE_REG, ("fpu_sqer:\n"));
DUMPFPN(FPE_REG, x);
DPRINTF(FPE_REG, ("=>\n"));
if (ISNAN(x)) {
fe->fe_cx |= FPSCR_VXSNAN;
DUMPFPN(FPE_REG, x);
return (x);
}
if (ISZERO(x)) {
fe->fe_cx |= FPSCR_ZX;
x->fp_class = FPC_INF;
DUMPFPN(FPE_REG, x);
return (x);
}
if (x->fp_sign) {
return (fpu_newnan(fe));
}
if (ISINF(x)) {
fe->fe_cx |= FPSCR_VXSQRT;
DUMPFPN(FPE_REG, 0);
return (0);
}
/*
* Calculate result exponent. As noted above, this may involve
* doubling the mantissa. We will also need to double x each
* time around the loop, so we define a macro for this here, and
* we break out the multiword mantissa.
*/
#ifdef FPU_SHL1_BY_ADD
#define DOUBLE_X { \
FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
}
#else
#define DOUBLE_X { \
x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
}
#endif
#if (FP_NMANT & 1) != 0
# define ODD_DOUBLE DOUBLE_X
# define EVEN_DOUBLE /* nothing */
#else
# define ODD_DOUBLE /* nothing */
# define EVEN_DOUBLE DOUBLE_X
#endif
x0 = x->fp_mant[0];
x1 = x->fp_mant[1];
x2 = x->fp_mant[2];
x3 = x->fp_mant[3];
e = x->fp_exp;
if (e & 1) /* exponent is odd; use sqrt(2mant) */
DOUBLE_X;
/* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */
/*
* Now calculate the mantissa root. Since x is now in [1..4),
* we know that the first trip around the loop will definitely
* set the top bit in q, so we can do that manually and start
* the loop at the next bit down instead. We must be sure to
* double x correctly while doing the `known q=1.0'.
*
* We do this one mantissa-word at a time, as noted above, to
* save work. To avoid `(1U << 31) << 1', we also do the top bit
* outside of each per-word loop.
*
* The calculation `t = y + bit' breaks down into `t0 = y0, ...,
* t3 = y3, t? |= bit' for the appropriate word. Since the bit
* is always a `new' one, this means that three of the `t?'s are
* just the corresponding `y?'; we use `#define's here for this.
* The variable `tt' holds the actual `t?' variable.
*/
/* calculate q0 */
#define t0 tt
bit = FP_1;
EVEN_DOUBLE;
/* if (x >= (t0 = y0 | bit)) { */ /* always true */
q = bit;
x0 -= bit;
y0 = bit << 1;
/* } */
ODD_DOUBLE;
while ((bit >>= 1) != 0) { /* for remaining bits in q0 */
EVEN_DOUBLE;
t0 = y0 | bit; /* t = y + bit */
if (x0 >= t0) { /* if x >= t then */
x0 -= t0; /* x -= t */
q |= bit; /* q += bit */
y0 |= bit << 1; /* y += bit << 1 */
}
ODD_DOUBLE;
}
x->fp_mant[0] = q;
#undef t0
/* calculate q1. note (y0&1)==0. */
#define t0 y0
#define t1 tt
q = 0;
y1 = 0;
bit = 1 << 31;
EVEN_DOUBLE;
t1 = bit;
FPU_SUBS(d1, x1, t1);
FPU_SUBC(d0, x0, t0); /* d = x - t */
if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */
x0 = d0, x1 = d1; /* x -= t */
q = bit; /* q += bit */
y0 |= 1; /* y += bit << 1 */
}
ODD_DOUBLE;
while ((bit >>= 1) != 0) { /* for remaining bits in q1 */
EVEN_DOUBLE; /* as before */
t1 = y1 | bit;
FPU_SUBS(d1, x1, t1);
FPU_SUBC(d0, x0, t0);
if ((int)d0 >= 0) {
x0 = d0, x1 = d1;
q |= bit;
y1 |= bit << 1;
}
ODD_DOUBLE;
}
x->fp_mant[1] = q;
#undef t1
/* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */
#define t1 y1
#define t2 tt
q = 0;
y2 = 0;
bit = 1 << 31;
EVEN_DOUBLE;
t2 = bit;
FPU_SUBS(d2, x2, t2);
FPU_SUBCS(d1, x1, t1);
FPU_SUBC(d0, x0, t0);
if ((int)d0 >= 0) {
x0 = d0, x1 = d1, x2 = d2;
q |= bit;
y1 |= 1; /* now t1, y1 are set in concrete */
}
ODD_DOUBLE;
while ((bit >>= 1) != 0) {
EVEN_DOUBLE;
t2 = y2 | bit;
FPU_SUBS(d2, x2, t2);
FPU_SUBCS(d1, x1, t1);
FPU_SUBC(d0, x0, t0);
if ((int)d0 >= 0) {
x0 = d0, x1 = d1, x2 = d2;
q |= bit;
y2 |= bit << 1;
}
ODD_DOUBLE;
}
x->fp_mant[2] = q;
#undef t2
/* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */
#define t2 y2
#define t3 tt
q = 0;
y3 = 0;
bit = 1 << 31;
EVEN_DOUBLE;
t3 = bit;
FPU_SUBS(d3, x3, t3);
FPU_SUBCS(d2, x2, t2);
FPU_SUBCS(d1, x1, t1);
FPU_SUBC(d0, x0, t0);
ODD_DOUBLE;
if ((int)d0 >= 0) {
x0 = d0, x1 = d1, x2 = d2;
q |= bit;
y2 |= 1;
}
while ((bit >>= 1) != 0) {
EVEN_DOUBLE;
t3 = y3 | bit;
FPU_SUBS(d3, x3, t3);
FPU_SUBCS(d2, x2, t2);
FPU_SUBCS(d1, x1, t1);
FPU_SUBC(d0, x0, t0);
if ((int)d0 >= 0) {
x0 = d0, x1 = d1, x2 = d2;
q |= bit;
y3 |= bit << 1;
}
ODD_DOUBLE;
}
x->fp_mant[3] = q;
/*
* The result, which includes guard and round bits, is exact iff
* x is now zero; any nonzero bits in x represent sticky bits.
*/
x->fp_sticky = x0 | x1 | x2 | x3;
DUMPFPN(FPE_REG, x);
return (x);
}
struct fpn *
fpu_div(struct fpemu *fe)
{
struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2;
u_int q, bit;
u_int r0, r1, r2, r3, d0, d1, d2, d3, y0, y1, y2, y3;
FPU_DECL_CARRY
/*
* Since divide is not commutative, we cannot just use ORDER.
* Check either operand for NaN first; if there is at least one,
* order the signalling one (if only one) onto the right, then
* return it. Otherwise we have the following cases:
*
* Inf / Inf = NaN, plus NV exception
* Inf / num = Inf [i.e., return x]
* Inf / 0 = Inf [i.e., return x]
* 0 / Inf = 0 [i.e., return x]
* 0 / num = 0 [i.e., return x]
* 0 / 0 = NaN, plus NV exception
* num / Inf = 0
* num / num = num (do the divide)
* num / 0 = Inf, plus DZ exception
*/
DPRINTF(FPE_REG, ("fpu_div:\n"));
DUMPFPN(FPE_REG, x);
DUMPFPN(FPE_REG, y);
DPRINTF(FPE_REG, ("=>\n"));
if (ISNAN(x) || ISNAN(y)) {
ORDER(x, y);
fe->fe_cx |= FPSCR_VXSNAN;
DUMPFPN(FPE_REG, y);
return (y);
}
/*
* Need to split the following out cause they generate different
* exceptions.
*/
if (ISINF(x)) {
if (x->fp_class == y->fp_class) {
fe->fe_cx |= FPSCR_VXIDI;
return (fpu_newnan(fe));
}
DUMPFPN(FPE_REG, x);
return (x);
}
if (ISZERO(x)) {
fe->fe_cx |= FPSCR_ZX;
if (x->fp_class == y->fp_class) {
fe->fe_cx |= FPSCR_VXZDZ;
return (fpu_newnan(fe));
}
DUMPFPN(FPE_REG, x);
return (x);
}
/* all results at this point use XOR of operand signs */
x->fp_sign ^= y->fp_sign;
if (ISINF(y)) {
x->fp_class = FPC_ZERO;
DUMPFPN(FPE_REG, x);
return (x);
}
if (ISZERO(y)) {
fe->fe_cx = FPSCR_ZX;
x->fp_class = FPC_INF;
DUMPFPN(FPE_REG, x);
return (x);
}
/*
* Macros for the divide. See comments at top for algorithm.
* Note that we expand R, D, and Y here.
*/
#define SUBTRACT /* D = R - Y */ \
FPU_SUBS(d3, r3, y3); FPU_SUBCS(d2, r2, y2); \
FPU_SUBCS(d1, r1, y1); FPU_SUBC(d0, r0, y0)
#define NONNEGATIVE /* D >= 0 */ \
((int)d0 >= 0)
#ifdef FPU_SHL1_BY_ADD
#define SHL1 /* R <<= 1 */ \
FPU_ADDS(r3, r3, r3); FPU_ADDCS(r2, r2, r2); \
FPU_ADDCS(r1, r1, r1); FPU_ADDC(r0, r0, r0)
#else
#define SHL1 \
r0 = (r0 << 1) | (r1 >> 31), r1 = (r1 << 1) | (r2 >> 31), \
r2 = (r2 << 1) | (r3 >> 31), r3 <<= 1
#endif
#define LOOP /* do ... while (bit >>= 1) */ \
do { \
SHL1; \
SUBTRACT; \
if (NONNEGATIVE) { \
q |= bit; \
r0 = d0, r1 = d1, r2 = d2, r3 = d3; \
} \
} while ((bit >>= 1) != 0)
#define WORD(r, i) /* calculate r->fp_mant[i] */ \
q = 0; \
bit = 1 << 31; \
LOOP; \
(x)->fp_mant[i] = q
/* Setup. Note that we put our result in x. */
r0 = x->fp_mant[0];
r1 = x->fp_mant[1];
r2 = x->fp_mant[2];
r3 = x->fp_mant[3];
y0 = y->fp_mant[0];
y1 = y->fp_mant[1];
y2 = y->fp_mant[2];
y3 = y->fp_mant[3];
bit = FP_1;
SUBTRACT;
if (NONNEGATIVE) {
x->fp_exp -= y->fp_exp;
r0 = d0, r1 = d1, r2 = d2, r3 = d3;
q = bit;
bit >>= 1;
} else {
/*
* The multiplication algorithm for normal numbers is as follows:
*
* The fraction of the product is built in the usual stepwise fashion.
* Each step consists of shifting the accumulator right one bit
* (maintaining any guard bits) and, if the next bit in y is set,
* adding the multiplicand (x) to the accumulator. Then, in any case,
* we advance one bit leftward in y. Algorithmically:
*
* A = 0;
* for (bit = 0; bit < FP_NMANT; bit++) {
* sticky |= A & 1, A >>= 1;
* if (Y & (1 << bit))
* A += X;
* }
*
* (X and Y here represent the mantissas of x and y respectively.)
* The resultant accumulator (A) is the product's mantissa. It may
* be as large as 11.11111... in binary and hence may need to be
* shifted right, but at most one bit.
*
* Since we do not have efficient multiword arithmetic, we code the
* accumulator as four separate words, just like any other mantissa.
* We use local variables in the hope that this is faster than memory.
* We keep x->fp_mant in locals for the same reason.
*
* In the algorithm above, the bits in y are inspected one at a time.
* We will pick them up 32 at a time and then deal with those 32, one
* at a time. Note, however, that we know several things about y:
*
* - the guard and round bits at the bottom are sure to be zero;
*
* - often many low bits are zero (y is often from a single or double
* precision source);
*
* - bit FP_NMANT-1 is set, and FP_1*2 fits in a word.
*
* We can also test for 32-zero-bits swiftly. In this case, the center
* part of the loop---setting sticky, shifting A, and not adding---will
* run 32 times without adding X to A. We can do a 32-bit shift faster
* by simply moving words. Since zeros are common, we optimize this case.
* Furthermore, since A is initially zero, we can omit the shift as well
* until we reach a nonzero word.
*/
struct fpn *
fpu_mul(struct fpemu *fe)
{
struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2;
u_int a3, a2, a1, a0, x3, x2, x1, x0, bit, m;
int sticky;
FPU_DECL_CARRY;
/*
* Put the `heavier' operand on the right (see fpu_emu.h).
* Then we will have one of the following cases, taken in the
* following order:
*
* - y = NaN. Implied: if only one is a signalling NaN, y is.
* The result is y.
* - y = Inf. Implied: x != NaN (is 0, number, or Inf: the NaN
* case was taken care of earlier).
* If x = 0, the result is NaN. Otherwise the result
* is y, with its sign reversed if x is negative.
* - x = 0. Implied: y is 0 or number.
* The result is 0 (with XORed sign as usual).
* - other. Implied: both x and y are numbers.
* The result is x * y (XOR sign, multiply bits, add exponents).
*/
DPRINTF(FPE_REG, ("fpu_mul:\n"));
DUMPFPN(FPE_REG, x);
DUMPFPN(FPE_REG, y);
DPRINTF(FPE_REG, ("=>\n"));
ORDER(x, y);
if (ISNAN(y)) {
y->fp_sign ^= x->fp_sign;
fe->fe_cx |= FPSCR_VXSNAN;
DUMPFPN(FPE_REG, y);
return (y);
}
if (ISINF(y)) {
if (ISZERO(x)) {
fe->fe_cx |= FPSCR_VXIMZ;
return (fpu_newnan(fe));
}
y->fp_sign ^= x->fp_sign;
DUMPFPN(FPE_REG, y);
return (y);
}
if (ISZERO(x)) {
x->fp_sign ^= y->fp_sign;
DUMPFPN(FPE_REG, x);
return (x);
}
/*
* Setup. In the code below, the mask `m' will hold the current
* mantissa byte from y. The variable `bit' denotes the bit
* within m. We also define some macros to deal with everything.
*/
x3 = x->fp_mant[3];
x2 = x->fp_mant[2];
x1 = x->fp_mant[1];
x0 = x->fp_mant[0];
sticky = a3 = a2 = a1 = a0 = 0;
#define ADD /* A += X */ \
FPU_ADDS(a3, a3, x3); \
FPU_ADDCS(a2, a2, x2); \
FPU_ADDCS(a1, a1, x1); \
FPU_ADDC(a0, a0, x0)
#define SHR1 /* A >>= 1, with sticky */ \
sticky |= a3 & 1, a3 = (a3 >> 1) | (a2 << 31), \
a2 = (a2 >> 1) | (a1 << 31), a1 = (a1 >> 1) | (a0 << 31), a0 >>= 1
#define SHR32 /* A >>= 32, with sticky */ \
sticky |= a3, a3 = a2, a2 = a1, a1 = a0, a0 = 0
#define STEP /* each 1-bit step of the multiplication */ \
SHR1; if (bit & m) { ADD; }; bit <<= 1
/*
* We are ready to begin. The multiply loop runs once for each
* of the four 32-bit words. Some words, however, are special.
* As noted above, the low order bits of Y are often zero. Even
* if not, the first loop can certainly skip the guard bits.
* The last word of y has its highest 1-bit in position FP_NMANT-1,
* so we stop the loop when we move past that bit.
*/
if ((m = y->fp_mant[3]) == 0) {
/* SHR32; */ /* unneeded since A==0 */
} else {
bit = 1 << FP_NG;
do {
STEP;
} while (bit != 0);
}
if ((m = y->fp_mant[2]) == 0) {
SHR32;
} else {
bit = 1;
do {
STEP;
} while (bit != 0);
}
if ((m = y->fp_mant[1]) == 0) {
SHR32;
} else {
bit = 1;
do {
STEP;
} while (bit != 0);
}
m = y->fp_mant[0]; /* definitely != 0 */
bit = 1;
do {
STEP;
} while (bit <= m);
/*
* Done with mantissa calculation. Get exponent and handle
* 11.111...1 case, then put result in place. We reuse x since
* it already has the right class (FP_NUM).
*/
m = x->fp_exp + y->fp_exp;
if (a0 >= FP_2) {
SHR1;
m++;
}
x->fp_sign ^= y->fp_sign;
x->fp_exp = m;
x->fp_sticky = sticky;
x->fp_mant[3] = a3;
x->fp_mant[2] = a2;
x->fp_mant[1] = a1;
x->fp_mant[0] = a0;
DUMPFPN(FPE_REG, x);
return (x);
}
struct fpn *
fpu_add(struct fpemu *fe)
{
struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2, *r;
u_int r0, r1, r2, r3;
int rd;
/*
* Put the `heavier' operand on the right (see fpu_emu.h).
* Then we will have one of the following cases, taken in the
* following order:
*
* - y = NaN. Implied: if only one is a signalling NaN, y is.
* The result is y.
* - y = Inf. Implied: x != NaN (is 0, number, or Inf: the NaN
* case was taken care of earlier).
* If x = -y, the result is NaN. Otherwise the result
* is y (an Inf of whichever sign).
* - y is 0. Implied: x = 0.
* If x and y differ in sign (one positive, one negative),
* the result is +0 except when rounding to -Inf. If same:
* +0 + +0 = +0; -0 + -0 = -0.
* - x is 0. Implied: y != 0.
* Result is y.
* - other. Implied: both x and y are numbers.
* Do addition a la Hennessey & Patterson.
*/
DPRINTF(FPE_REG, ("fpu_add:\n"));
DUMPFPN(FPE_REG, x);
DUMPFPN(FPE_REG, y);
DPRINTF(FPE_REG, ("=>\n"));
ORDER(x, y);
if (ISNAN(y)) {
fe->fe_cx |= FPSCR_VXSNAN;
DUMPFPN(FPE_REG, y);
return (y);
}
if (ISINF(y)) {
if (ISINF(x) && x->fp_sign != y->fp_sign) {
fe->fe_cx |= FPSCR_VXISI;
return (fpu_newnan(fe));
}
DUMPFPN(FPE_REG, y);
return (y);
}
rd = ((fe->fe_fpscr) & FPSCR_RN);
if (ISZERO(y)) {
if (rd != FSR_RD_RM) /* only -0 + -0 gives -0 */
y->fp_sign &= x->fp_sign;
else /* any -0 operand gives -0 */
y->fp_sign |= x->fp_sign;
DUMPFPN(FPE_REG, y);
return (y);
}
if (ISZERO(x)) {
DUMPFPN(FPE_REG, y);
return (y);
}
/*
* We really have two numbers to add, although their signs may
* differ. Make the exponents match, by shifting the smaller
* number right (e.g., 1.011 => 0.1011) and increasing its
* exponent (2^3 => 2^4). Note that we do not alter the exponents
* of x and y here.
*/
r = &fe->fe_f3;
r->fp_class = FPC_NUM;
if (x->fp_exp == y->fp_exp) {
r->fp_exp = x->fp_exp;
r->fp_sticky = 0;
} else {
if (x->fp_exp < y->fp_exp) {
/*
* Try to avoid subtract case iii (see below).
* This also guarantees that x->fp_sticky = 0.
*/
SWAP(x, y);
}
/* now x->fp_exp > y->fp_exp */
r->fp_exp = x->fp_exp;
r->fp_sticky = fpu_shr(y, x->fp_exp - y->fp_exp);
}
r->fp_sign = x->fp_sign;
if (x->fp_sign == y->fp_sign) {
FPU_DECL_CARRY
/*
* The signs match, so we simply add the numbers. The result
* may be `supernormal' (as big as 1.111...1 + 1.111...1, or
* 11.111...0). If so, a single bit shift-right will fix it
* (but remember to adjust the exponent).
*/
/* r->fp_mant = x->fp_mant + y->fp_mant */
FPU_ADDS(r->fp_mant[3], x->fp_mant[3], y->fp_mant[3]);
FPU_ADDCS(r->fp_mant[2], x->fp_mant[2], y->fp_mant[2]);
FPU_ADDCS(r->fp_mant[1], x->fp_mant[1], y->fp_mant[1]);
FPU_ADDC(r0, x->fp_mant[0], y->fp_mant[0]);
if ((r->fp_mant[0] = r0) >= FP_2) {
(void) fpu_shr(r, 1);
r->fp_exp++;
}
} else {
FPU_DECL_CARRY
/*
* The signs differ, so things are rather more difficult.
* H&P would have us negate the negative operand and add;
* this is the same as subtracting the negative operand.
* This is quite a headache. Instead, we will subtract
* y from x, regardless of whether y itself is the negative
* operand. When this is done one of three conditions will
* hold, depending on the magnitudes of x and y:
* case i) |x| > |y|. The result is just x - y,
* with x's sign, but it may need to be normalized.
* case ii) |x| = |y|. The result is 0 (maybe -0)
* so must be fixed up.
* case iii) |x| < |y|. We goofed; the result should
* be (y - x), with the same sign as y.
* We could compare |x| and |y| here and avoid case iii,
* but that would take just as much work as the subtract.
* We can tell case iii has occurred by an overflow.
*
* N.B.: since x->fp_exp >= y->fp_exp, x->fp_sticky = 0.
*/
/* r->fp_mant = x->fp_mant - y->fp_mant */
FPU_SET_CARRY(y->fp_sticky);
FPU_SUBCS(r3, x->fp_mant[3], y->fp_mant[3]);
FPU_SUBCS(r2, x->fp_mant[2], y->fp_mant[2]);
FPU_SUBCS(r1, x->fp_mant[1], y->fp_mant[1]);
FPU_SUBC(r0, x->fp_mant[0], y->fp_mant[0]);
if (r0 < FP_2) {
/* cases i and ii */
if ((r0 | r1 | r2 | r3) == 0) {
/* case ii */
r->fp_class = FPC_ZERO;
r->fp_sign = rd == FSR_RD_RM;
return (r);
}
} else {
/*
* Oops, case iii. This can only occur when the
* exponents were equal, in which case neither
* x nor y have sticky bits set. Flip the sign
* (to y's sign) and negate the result to get y - x.
*/
#ifdef DIAGNOSTIC
if (x->fp_exp != y->fp_exp || r->fp_sticky)
panic("fpu_add");
#endif
r->fp_sign = y->fp_sign;
FPU_SUBS(r3, 0, r3);
FPU_SUBCS(r2, 0, r2);
FPU_SUBCS(r1, 0, r1);
FPU_SUBC(r0, 0, r0);
}
r->fp_mant[3] = r3;
r->fp_mant[2] = r2;
r->fp_mant[1] = r1;
r->fp_mant[0] = r0;
if (r0 < FP_1)
fpu_norm(r);
}
DUMPFPN(FPE_REG, r);
return (r);
}
static struct fpn *
__fpu_modrem(struct fpemu *fe, int is_mod)
{
static struct fpn X, Y;
struct fpn *x, *y, *r;
uint32_t signX, signY, signQ;
int j, k, l, q;
int cmp;
if (ISNAN(&fe->fe_f1) || ISNAN(&fe->fe_f2))
return fpu_newnan(fe);
if (ISINF(&fe->fe_f1) || ISZERO(&fe->fe_f2))
return fpu_newnan(fe);
CPYFPN(&X, &fe->fe_f1);
CPYFPN(&Y, &fe->fe_f2);
x = &X;
y = &Y;
q = 0;
r = &fe->fe_f2;
/*
* Step 1
*/
signX = x->fp_sign;
signY = y->fp_sign;
signQ = (signX ^ signY);
x->fp_sign = y->fp_sign = 0;
/* Special treatment that just return input value but Q is necessary */
if (ISZERO(x) || ISINF(y)) {
r = &fe->fe_f1;
goto Step7;
}
/*
* Step 2
*/
l = x->fp_exp - y->fp_exp;
k = 0;
CPYFPN(r, x);
if (l >= 0) {
r->fp_exp -= l;
j = l;
/*
* Step 3
*/
for (;;) {
cmp = abscmp3(r, y);
/* Step 3.1 */
if (cmp == 0)
break;
/* Step 3.2 */
if (cmp > 0) {
CPYFPN(&fe->fe_f1, r);
CPYFPN(&fe->fe_f2, y);
fe->fe_f2.fp_sign = 1;
r = fpu_add(fe);
q++;
}
/* Step 3.3 */
if (j == 0)
goto Step4;
/* Step 3.4 */
k++;
j--;
q += q;
r->fp_exp++;
}
/* R == Y */
q++;
r->fp_class = FPC_ZERO;
goto Step7;
}
Step4:
r->fp_sign = signX;
/*
* Step 5
*/
if (is_mod)
goto Step7;
/*
* Step 6
*/
/* y = y / 2 */
y->fp_exp--;
/* abscmp3 ignore sign */
cmp = abscmp3(r, y);
/* revert y */
y->fp_exp++;
if (cmp > 0 || (cmp == 0 && q % 2)) {
q++;
CPYFPN(&fe->fe_f1, r);
CPYFPN(&fe->fe_f2, y);
fe->fe_f2.fp_sign = !signX;
r = fpu_add(fe);
}
/*
* Step 7
*/
Step7:
q &= 0x7f;
q |= (signQ << 7);
fe->fe_fpframe->fpf_fpsr =
fe->fe_fpsr =
(fe->fe_fpsr & ~FPSR_QTT) | (q << 16);
return r;
}
/*
* sin(x):
*
* if (x < 0) {
* x = abs(x);
* sign = 1;
* }
* if (x > 2*pi) {
* x %= 2*pi;
* }
* if (x > pi) {
* x -= pi;
* sign inverse;
* }
* if (x > pi/2) {
* y = cos(x - pi/2);
* } else {
* y = sin(x);
* }
* if (sign) {
* y = -y;
* }
*/
struct fpn *
fpu_sin(struct fpemu *fe)
{
struct fpn x;
struct fpn p;
struct fpn *r;
int sign;
if (ISNAN(&fe->fe_f2))
return &fe->fe_f2;
if (ISINF(&fe->fe_f2))
return fpu_newnan(fe);
/* if x is +0/-0, return +0/-0 */
if (ISZERO(&fe->fe_f2))
return &fe->fe_f2;
CPYFPN(&x, &fe->fe_f2);
/* x = abs(input) */
sign = x.fp_sign;
x.fp_sign = 0;
/* p <- 2*pi */
fpu_const(&p, FPU_CONST_PI);
p.fp_exp++;
/*
* if (x > 2*pi*N)
* sin(x) is sin(x - 2*pi*N)
*/
CPYFPN(&fe->fe_f1, &x);
CPYFPN(&fe->fe_f2, &p);
r = fpu_cmp(fe);
if (r->fp_sign == 0) {
CPYFPN(&fe->fe_f1, &x);
CPYFPN(&fe->fe_f2, &p);
r = fpu_mod(fe);
CPYFPN(&x, r);
}
/* p <- pi */
p.fp_exp--;
/*
* if (x > pi)
* sin(x) is -sin(x - pi)
*/
CPYFPN(&fe->fe_f1, &x);
CPYFPN(&fe->fe_f2, &p);
fe->fe_f2.fp_sign = 1;
r = fpu_add(fe);
if (r->fp_sign == 0) {
CPYFPN(&x, r);
sign ^= 1;
}
/* p <- pi/2 */
p.fp_exp--;
/*
* if (x > pi/2)
* sin(x) is cos(x - pi/2)
* else
* sin(x)
*/
CPYFPN(&fe->fe_f1, &x);
CPYFPN(&fe->fe_f2, &p);
fe->fe_f2.fp_sign = 1;
r = fpu_add(fe);
if (r->fp_sign == 0) {
__fpu_sincos_cordic(fe, r);
r = &fe->fe_f2;
} else {
__fpu_sincos_cordic(fe, &x);
r = &fe->fe_f1;
}
r->fp_sign = sign;
return r;
}
