8302040: Port fdlibm sqrt to Java

Reviewed-by: bpb, thartmann, aturbanov
This commit is contained in:
Joe Darcy 2023-02-28 18:33:53 +00:00
parent 6423065b7d
commit 61e8867591
8 changed files with 875 additions and 6 deletions
src
hotspot/share
java.base/share/classes/java/lang
test/jdk/java/lang

@ -202,8 +202,8 @@ class methodHandle;
do_intrinsic(_maxF_strict, java_lang_StrictMath, max_name, float2_float_signature, F_S) \
do_intrinsic(_minD_strict, java_lang_StrictMath, min_name, double2_double_signature, F_S) \
do_intrinsic(_maxD_strict, java_lang_StrictMath, max_name, double2_double_signature, F_S) \
/* Special flavor of dsqrt intrinsic to handle the "native" method in StrictMath. Otherwise the same as in Math. */ \
do_intrinsic(_dsqrt_strict, java_lang_StrictMath, sqrt_name, double_double_signature, F_SN) \
/* Additional dsqrt intrinsic to directly handle the sqrt method in StrictMath. Otherwise the same as in Math. */ \
do_intrinsic(_dsqrt_strict, java_lang_StrictMath, sqrt_name, double_double_signature, F_S) \
\
do_intrinsic(_floatIsInfinite, java_lang_Float, isInfinite_name, float_bool_signature, F_S) \
do_name( isInfinite_name, "isInfinite") \

@ -148,7 +148,7 @@ AbstractInterpreter::MethodKind AbstractInterpreter::method_kind(const methodHan
case vmIntrinsics::_fmaD: return java_lang_math_fmaD;
case vmIntrinsics::_fmaF: return java_lang_math_fmaF;
case vmIntrinsics::_dsqrt: return java_lang_math_sqrt;
case vmIntrinsics::_dsqrt_strict: return native;
case vmIntrinsics::_dsqrt_strict: return java_lang_math_sqrt;
case vmIntrinsics::_Reference_get: return java_lang_ref_reference_get;
case vmIntrinsics::_Object_init:
if (RegisterFinalizersAtInit && m->code_size() == 1) {

@ -104,6 +104,15 @@ class FdLibm {
( ((long)high)) << 32 );
}
/**
* Return a double with its high-order bits of the first argument
* and the low-order bits of the second argument..
*/
private static double __HI_LO(int high, int low) {
return Double.longBitsToDouble(((long)high << 32) |
(low & 0xffff_ffffL));
}
/** Returns the arcsine of x.
*
* Method :
@ -504,6 +513,449 @@ class FdLibm {
}
}
/**
* Return correctly rounded sqrt.
* ------------------------------------------
* | Use the hardware sqrt if you have one |
* ------------------------------------------
* Method:
* Bit by bit method using integer arithmetic. (Slow, but portable)
* 1. Normalization
* Scale x to y in [1,4) with even powers of 2:
* find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
* sqrt(x) = 2^k * sqrt(y)
* 2. Bit by bit computation
* Let q = sqrt(y) truncated to i bit after binary point (q = 1),
* i 0
* i+1 2
* s = 2*q , and y = 2 * ( y - q ). (1)
* i i i i
*
* To compute q from q , one checks whether
* i+1 i
*
* -(i+1) 2
* (q + 2 ) <= y. (2)
* i
* -(i+1)
* If (2) is false, then q = q ; otherwise q = q + 2 .
* i+1 i i+1 i
*
* With some algebraic manipulation, it is not difficult to see
* that (2) is equivalent to
* -(i+1)
* s + 2 <= y (3)
* i i
*
* The advantage of (3) is that s and y can be computed by
* i i
* the following recurrence formula:
* if (3) is false
*
* s = s , y = y ; (4)
* i+1 i i+1 i
*
* otherwise,
* -i -(i+1)
* s = s + 2 , y = y - s - 2 (5)
* i+1 i i+1 i i
*
* One may easily use induction to prove (4) and (5).
* Note. Since the left hand side of (3) contain only i+2 bits,
* it does not necessary to do a full (53-bit) comparison
* in (3).
* 3. Final rounding
* After generating the 53 bits result, we compute one more bit.
* Together with the remainder, we can decide whether the
* result is exact, bigger than 1/2ulp, or less than 1/2ulp
* (it will never equal to 1/2ulp).
* The rounding mode can be detected by checking whether
* huge + tiny is equal to huge, and whether huge - tiny is
* equal to huge for some floating point number "huge" and "tiny".
*
* Special cases:
* sqrt(+-0) = +-0 ... exact
* sqrt(inf) = inf
* sqrt(-ve) = NaN ... with invalid signal
* sqrt(NaN) = NaN ... with invalid signal for signaling NaN
*
* Other methods : see the appended file at the end of the program below.
*---------------
*/
static class Sqrt {
private Sqrt() {throw new UnsupportedOperationException();}
private static final double tiny = 1.0e-300;
static double compute(double x) {
double z = 0.0;
int sign = 0x8000_0000;
/*unsigned*/ int r, t1, s1, ix1, q1;
int ix0, s0, q, m, t, i;
ix0 = __HI(x); // high word of x
ix1 = __LO(x); // low word of x
// take care of Inf and NaN
if ((ix0 & 0x7ff0_0000) == 0x7ff0_0000) {
return x*x + x; // sqrt(NaN)=NaN, sqrt(+inf)=+inf, sqrt(-inf)=sNaN
}
// take care of zero
if (ix0 <= 0) {
if (((ix0 & (~sign)) | ix1) == 0)
return x; // sqrt(+-0) = +-0
else if (ix0 < 0)
return (x-x)/(x-x); // sqrt(-ve) = sNaN
}
// normalize x
m = (ix0 >> 20);
if (m == 0) { // subnormal x
while (ix0 == 0) {
m -= 21;
ix0 |= (ix1 >>> 11); // unsigned shift
ix1 <<= 21;
}
for(i = 0; (ix0 & 0x0010_0000) == 0; i++) {
ix0 <<= 1;
}
m -= i-1;
ix0 |= (ix1 >>> (32 - i)); // unsigned shift
ix1 <<= i;
}
m -= 1023; // unbias exponent */
ix0 = (ix0 & 0x000f_ffff) | 0x0010_0000;
if ((m & 1) != 0){ // odd m, double x to make it even
ix0 += ix0 + ((ix1 & sign) >>> 31); // unsigned shift
ix1 += ix1;
}
m >>= 1; // m = [m/2]
// generate sqrt(x) bit by bit
ix0 += ix0 + ((ix1 & sign) >>> 31); // unsigned shift
ix1 += ix1;
q = q1 = s0 = s1 = 0; // [q,q1] = sqrt(x)
r = 0x0020_0000; // r = moving bit from right to left
while (r != 0) {
t = s0 + r;
if (t <= ix0) {
s0 = t + r;
ix0 -= t;
q += r;
}
ix0 += ix0 + ((ix1 & sign) >>> 31); // unsigned shift
ix1 += ix1;
r >>>= 1; // unsigned shift
}
r = sign;
while (r != 0) {
t1 = s1 + r;
t = s0;
if ((t < ix0) ||
((t == ix0) && (Integer.compareUnsigned(t1, ix1) <= 0 ))) { // t1 <= ix1
s1 = t1 + r;
if (((t1 & sign) == sign) && (s1 & sign) == 0) {
s0 += 1;
}
ix0 -= t;
if (Integer.compareUnsigned(ix1, t1) < 0) { // ix1 < t1
ix0 -= 1;
}
ix1 -= t1;
q1 += r;
}
ix0 += ix0 + ((ix1 & sign) >>> 31); // unsigned shift
ix1 += ix1;
r >>>= 1; // unsigned shift
}
// use floating add to find out rounding direction
if ((ix0 | ix1) != 0) {
z = 1.0 - tiny; // trigger inexact flag
if (z >= 1.0) {
z = 1.0 + tiny;
if (q1 == 0xffff_ffff) {
q1 = 0;
q += 1;
} else if (z > 1.0) {
if (q1 == 0xffff_fffe) {
q += 1;
}
q1 += 2;
} else {
q1 += (q1 & 1);
}
}
}
ix0 = (q >> 1) + 0x3fe0_0000;
ix1 = q1 >>> 1; // unsigned shift
if ((q & 1) == 1) {
ix1 |= sign;
}
ix0 += (m << 20);
return __HI_LO(ix0, ix1);
}
}
// The following comment is supplementary information from the FDLIBM sources.
/*
* Other methods (use floating-point arithmetic)
* -------------
* (This is a copy of a drafted paper by Prof W. Kahan
* and K.C. Ng, written in May, 1986)
*
* Two algorithms are given here to implement sqrt(x)
* (IEEE double precision arithmetic) in software.
* Both supply sqrt(x) correctly rounded. The first algorithm (in
* Section A) uses newton iterations and involves four divisions.
* The second one uses reciproot iterations to avoid division, but
* requires more multiplications. Both algorithms need the ability
* to chop results of arithmetic operations instead of round them,
* and the INEXACT flag to indicate when an arithmetic operation
* is executed exactly with no roundoff error, all part of the
* standard (IEEE 754-1985). The ability to perform shift, add,
* subtract and logical AND operations upon 32-bit words is needed
* too, though not part of the standard.
*
* A. sqrt(x) by Newton Iteration
*
* (1) Initial approximation
*
* Let x0 and x1 be the leading and the trailing 32-bit words of
* a floating point number x (in IEEE double format) respectively
*
* 1 11 52 ...widths
* ------------------------------------------------------
* x: |s| e | f |
* ------------------------------------------------------
* msb lsb msb lsb ...order
*
*
* ------------------------ ------------------------
* x0: |s| e | f1 | x1: | f2 |
* ------------------------ ------------------------
*
* By performing shifts and subtracts on x0 and x1 (both regarded
* as integers), we obtain an 8-bit approximation of sqrt(x) as
* follows.
*
* k := (x0>>1) + 0x1ff80000;
* y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits
* Here k is a 32-bit integer and T1[] is an integer array containing
* correction terms. Now magically the floating value of y (y's
* leading 32-bit word is y0, the value of its trailing word is 0)
* approximates sqrt(x) to almost 8-bit.
*
* Value of T1:
* static int T1[32]= {
* 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592,
* 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215,
* 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581,
* 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,};
*
* (2) Iterative refinement
*
* Apply Heron's rule three times to y, we have y approximates
* sqrt(x) to within 1 ulp (Unit in the Last Place):
*
* y := (y+x/y)/2 ... almost 17 sig. bits
* y := (y+x/y)/2 ... almost 35 sig. bits
* y := y-(y-x/y)/2 ... within 1 ulp
*
*
* Remark 1.
* Another way to improve y to within 1 ulp is:
*
* y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x)
* y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x)
*
* 2
* (x-y )*y
* y := y + 2* ---------- ...within 1 ulp
* 2
* 3y + x
*
*
* This formula has one division fewer than the one above; however,
* it requires more multiplications and additions. Also x must be
* scaled in advance to avoid spurious overflow in evaluating the
* expression 3y*y+x. Hence it is not recommended uless division
* is slow. If division is very slow, then one should use the
* reciproot algorithm given in section B.
*
* (3) Final adjustment
*
* By twiddling y's last bit it is possible to force y to be
* correctly rounded according to the prevailing rounding mode
* as follows. Let r and i be copies of the rounding mode and
* inexact flag before entering the square root program. Also we
* use the expression y+-ulp for the next representable floating
* numbers (up and down) of y. Note that y+-ulp = either fixed
* point y+-1, or multiply y by nextafter(1,+-inf) in chopped
* mode.
*
* I := FALSE; ... reset INEXACT flag I
* R := RZ; ... set rounding mode to round-toward-zero
* z := x/y; ... chopped quotient, possibly inexact
* If(not I) then { ... if the quotient is exact
* if(z=y) {
* I := i; ... restore inexact flag
* R := r; ... restore rounded mode
* return sqrt(x):=y.
* } else {
* z := z - ulp; ... special rounding
* }
* }
* i := TRUE; ... sqrt(x) is inexact
* If (r=RN) then z=z+ulp ... rounded-to-nearest
* If (r=RP) then { ... round-toward-+inf
* y = y+ulp; z=z+ulp;
* }
* y := y+z; ... chopped sum
* y0:=y0-0x00100000; ... y := y/2 is correctly rounded.
* I := i; ... restore inexact flag
* R := r; ... restore rounded mode
* return sqrt(x):=y.
*
* (4) Special cases
*
* Square root of +inf, +-0, or NaN is itself;
* Square root of a negative number is NaN with invalid signal.
*
*
* B. sqrt(x) by Reciproot Iteration
*
* (1) Initial approximation
*
* Let x0 and x1 be the leading and the trailing 32-bit words of
* a floating point number x (in IEEE double format) respectively
* (see section A). By performing shifs and subtracts on x0 and y0,
* we obtain a 7.8-bit approximation of 1/sqrt(x) as follows.
*
* k := 0x5fe80000 - (x0>>1);
* y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits
*
* Here k is a 32-bit integer and T2[] is an integer array
* containing correction terms. Now magically the floating
* value of y (y's leading 32-bit word is y0, the value of
* its trailing word y1 is set to zero) approximates 1/sqrt(x)
* to almost 7.8-bit.
*
* Value of T2:
* static int T2[64]= {
* 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866,
* 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f,
* 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d,
* 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0,
* 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989,
* 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd,
* 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e,
* 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,};
*
* (2) Iterative refinement
*
* Apply Reciproot iteration three times to y and multiply the
* result by x to get an approximation z that matches sqrt(x)
* to about 1 ulp. To be exact, we will have
* -1ulp < sqrt(x)-z<1.0625ulp.
*
* ... set rounding mode to Round-to-nearest
* y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x)
* y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x)
* ... special arrangement for better accuracy
* z := x*y ... 29 bits to sqrt(x), with z*y<1
* z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x)
*
* Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that
* (a) the term z*y in the final iteration is always less than 1;
* (b) the error in the final result is biased upward so that
* -1 ulp < sqrt(x) - z < 1.0625 ulp
* instead of |sqrt(x)-z|<1.03125ulp.
*
* (3) Final adjustment
*
* By twiddling y's last bit it is possible to force y to be
* correctly rounded according to the prevailing rounding mode
* as follows. Let r and i be copies of the rounding mode and
* inexact flag before entering the square root program. Also we
* use the expression y+-ulp for the next representable floating
* numbers (up and down) of y. Note that y+-ulp = either fixed
* point y+-1, or multiply y by nextafter(1,+-inf) in chopped
* mode.
*
* R := RZ; ... set rounding mode to round-toward-zero
* switch(r) {
* case RN: ... round-to-nearest
* if(x<= z*(z-ulp)...chopped) z = z - ulp; else
* if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp;
* break;
* case RZ:case RM: ... round-to-zero or round-to--inf
* R:=RP; ... reset rounding mod to round-to-+inf
* if(x<z*z ... rounded up) z = z - ulp; else
* if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp;
* break;
* case RP: ... round-to-+inf
* if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else
* if(x>z*z ...chopped) z = z+ulp;
* break;
* }
*
* Remark 3. The above comparisons can be done in fixed point. For
* example, to compare x and w=z*z chopped, it suffices to compare
* x1 and w1 (the trailing parts of x and w), regarding them as
* two's complement integers.
*
* ...Is z an exact square root?
* To determine whether z is an exact square root of x, let z1 be the
* trailing part of z, and also let x0 and x1 be the leading and
* trailing parts of x.
*
* If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0
* I := 1; ... Raise Inexact flag: z is not exact
* else {
* j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2
* k := z1 >> 26; ... get z's 25-th and 26-th
* fraction bits
* I := i or (k&j) or ((k&(j+j+1))!=(x1&3));
* }
* R:= r ... restore rounded mode
* return sqrt(x):=z.
*
* If multiplication is cheaper then the foregoing red tape, the
* Inexact flag can be evaluated by
*
* I := i;
* I := (z*z!=x) or I.
*
* Note that z*z can overwrite I; this value must be sensed if it is
* True.
*
* Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be
* zero.
*
* --------------------
* z1: | f2 |
* --------------------
* bit 31 bit 0
*
* Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd
* or even of logb(x) have the following relations:
*
* -------------------------------------------------
* bit 27,26 of z1 bit 1,0 of x1 logb(x)
* -------------------------------------------------
* 00 00 odd and even
* 01 01 even
* 10 10 odd
* 10 00 even
* 11 01 even
* -------------------------------------------------
*
* (4) Special cases (see (4) of Section A).
*/
/**
* cbrt(x)
* Return cube root of x

@ -310,7 +310,9 @@ public final class StrictMath {
* @return the positive square root of {@code a}.
*/
@IntrinsicCandidate
public static native double sqrt(double a);
public static double sqrt(double a) {
return FdLibm.Sqrt.compute(a);
}
/**
* Returns the cube root of a {@code double} value. For

@ -0,0 +1,105 @@
/*
* Copyright (c) 2003, 2023, Oracle and/or its affiliates. All rights reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
* or visit www.oracle.com if you need additional information or have any
* questions.
*/
/*
* @test
* @library /test/lib
* @build Tests
* @run main SqrtTests
* @bug 8302040
* @summary Tests for {Math, StrictMath}.sqrt
*/
public class SqrtTests {
private SqrtTests(){}
public static void main(String... argv) {
int failures = 0;
failures += testSqrt();
if (failures > 0) {
System.err.println("Testing sqrt incurred "
+ failures + " failures.");
throw new RuntimeException();
}
}
private static final double InfinityD = Double.POSITIVE_INFINITY;
private static final double NaNd = Double.NaN;
/**
* "Returns the correctly rounded positive square root of a double value. Special cases:
*
* If the argument is NaN or less than zero, then the result is NaN.
*
* If the argument is positive infinity, then the result is positive infinity.
*
* If the argument is positive zero or negative zero, then the
* result is the same as the argument.
*
* Otherwise, the result is the double value closest to the true
* mathematical square root of the argument value."
*/
private static int testSqrt() {
int failures = 0;
for(double nan : Tests.NaNs) {
failures += testSqrtCase(nan, NaNd);
}
double [][] testCases = {
{InfinityD, InfinityD},
{-Double.MIN_VALUE, NaNd},
{-Double.MIN_NORMAL, NaNd},
{-Double.MAX_VALUE, NaNd},
{-InfinityD, NaNd},
{+0.0, +0.0},
{-0.0, -0.0},
// Test some notable perfect squares
{+0.25, +0.5},
{+1.0, +1.0},
{+4.0, +2.0},
{+9.0, +3.0},
{+0x1.ffffff0000002p1023, +0x1.ffffff8p511}
};
for(int i = 0; i < testCases.length; i++) {
failures += testSqrtCase(testCases[i][0], testCases[i][1]);
}
return failures;
}
private static int testSqrtCase(double input, double expected) {
int failures=0;
failures+=Tests.test("Math.sqrt", input, Math::sqrt, expected);
failures+=Tests.test("StrictMath.sqrt", input, StrictMath::sqrt, expected);
return failures;
}
}

@ -23,7 +23,7 @@
/*
* @test
* @bug 8301833 8302026 8301444 8302028
* @bug 8301833 8302026 8301444 8302028 8302040
* @build Tests
* @build FdlibmTranslit
* @build ExhaustingTests
@ -66,7 +66,11 @@ public class ExhaustingTests {
private static long testUnaryMethods() {
long failures = 0;
UnaryTestCase[] testCases = {
// new UnaryTestCase("sqrt", FdlibmTranslit::sqrt, StrictMath::sqrt, DEFAULT_SHIFT),
// Since sqrt is correctly rounded and thus for each input
// there is one well-defined correct result, additional
// comparison of the transliteration sqrt or StrictMath
// sqrt could be made against Math::sqrt.
new UnaryTestCase("sqrt", FdlibmTranslit::sqrt, StrictMath::sqrt, DEFAULT_SHIFT),
new UnaryTestCase("cbrt", FdlibmTranslit::cbrt, StrictMath::cbrt, DEFAULT_SHIFT),
new UnaryTestCase("log", FdlibmTranslit::log, StrictMath::log, DEFAULT_SHIFT),

@ -90,6 +90,10 @@ public class FdlibmTranslit {
return Hypot.compute(x, y);
}
public static double sqrt(double x) {
return Sqrt.compute(x);
}
public static double cbrt(double x) {
return Cbrt.compute(x);
}
@ -504,6 +508,178 @@ public class FdlibmTranslit {
}
}
/**
* Return correctly rounded sqrt.
* ------------------------------------------
* | Use the hardware sqrt if you have one |
* ------------------------------------------
* Method:
* Bit by bit method using integer arithmetic. (Slow, but portable)
* 1. Normalization
* Scale x to y in [1,4) with even powers of 2:
* find an integer k such that 1 <= (y=x*2^(2k)) < 4, then
* sqrt(x) = 2^k * sqrt(y)
* 2. Bit by bit computation
* Let q = sqrt(y) truncated to i bit after binary point (q = 1),
* i 0
* i+1 2
* s = 2*q , and y = 2 * ( y - q ). (1)
* i i i i
*
* To compute q from q , one checks whether
* i+1 i
*
* -(i+1) 2
* (q + 2 ) <= y. (2)
* i
* -(i+1)
* If (2) is false, then q = q ; otherwise q = q + 2 .
* i+1 i i+1 i
*
* With some algebraic manipulation, it is not difficult to see
* that (2) is equivalent to
* -(i+1)
* s + 2 <= y (3)
* i i
*
* The advantage of (3) is that s and y can be computed by
* i i
* the following recurrence formula:
* if (3) is false
*
* s = s , y = y ; (4)
* i+1 i i+1 i
*
* otherwise,
* -i -(i+1)
* s = s + 2 , y = y - s - 2 (5)
* i+1 i i+1 i i
*
* One may easily use induction to prove (4) and (5).
* Note. Since the left hand side of (3) contain only i+2 bits,
* it does not necessary to do a full (53-bit) comparison
* in (3).
* 3. Final rounding
* After generating the 53 bits result, we compute one more bit.
* Together with the remainder, we can decide whether the
* result is exact, bigger than 1/2ulp, or less than 1/2ulp
* (it will never equal to 1/2ulp).
* The rounding mode can be detected by checking whether
* huge + tiny is equal to huge, and whether huge - tiny is
* equal to huge for some floating point number "huge" and "tiny".
*
* Special cases:
* sqrt(+-0) = +-0 ... exact
* sqrt(inf) = inf
* sqrt(-ve) = NaN ... with invalid signal
* sqrt(NaN) = NaN ... with invalid signal for signaling NaN
*
* Other methods : see the appended file at the end of the program below.
*---------------
*/
static class Sqrt {
private static final double one = 1.0, tiny=1.0e-300;
public static double compute(double x) {
double z = 0.0;
int sign = (int)0x80000000;
/*unsigned*/ int r,t1,s1,ix1,q1;
int ix0,s0,q,m,t,i;
ix0 = __HI(x); /* high word of x */
ix1 = __LO(x); /* low word of x */
/* take care of Inf and NaN */
if((ix0&0x7ff00000)==0x7ff00000) {
return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf
sqrt(-inf)=sNaN */
}
/* take care of zero */
if(ix0<=0) {
if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */
else if(ix0<0)
return (x-x)/(x-x); /* sqrt(-ve) = sNaN */
}
/* normalize x */
m = (ix0>>20);
if(m==0) { /* subnormal x */
while(ix0==0) {
m -= 21;
ix0 |= (ix1>>>11); ix1 <<= 21; // unsigned shift
}
for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1;
m -= i-1;
ix0 |= (ix1>>>(32-i)); // unsigned shift
ix1 <<= i;
}
m -= 1023; /* unbias exponent */
ix0 = (ix0&0x000fffff)|0x00100000;
if((m&1) != 0){ /* odd m, double x to make it even */
ix0 += ix0 + ((ix1&sign)>>>31); // unsigned shift
ix1 += ix1;
}
m >>= 1; /* m = [m/2] */
/* generate sqrt(x) bit by bit */
ix0 += ix0 + ((ix1&sign)>>>31); // unsigned shift
ix1 += ix1;
q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
r = 0x00200000; /* r = moving bit from right to left */
while(r!=0) {
t = s0+r;
if(t<=ix0) {
s0 = t+r;
ix0 -= t;
q += r;
}
ix0 += ix0 + ((ix1&sign)>>>31); // unsigned shift
ix1 += ix1;
r>>>=1; // unsigned shift
}
r = sign;
while(r!=0) {
t1 = s1+r;
t = s0;
if((t<ix0)||((t==ix0)&&(Integer.compareUnsigned(t1, ix1) <= 0 ))) { // t1<=ix1
s1 = t1+r;
if(((t1&sign)==sign)&&(s1&sign)==0) s0 += 1;
ix0 -= t;
if (Integer.compareUnsigned(ix1, t1) < 0) ix0 -= 1; // ix1 < t1
ix1 -= t1;
q1 += r;
}
ix0 += ix0 + ((ix1&sign)>>>31); // unsigned shift
ix1 += ix1;
r>>>=1; // unsigned shift
}
/* use floating add to find out rounding direction */
if((ix0|ix1)!=0) {
z = one-tiny; /* trigger inexact flag */
if (z>=one) {
z = one+tiny;
if (q1==0xffffffff) { q1=0; q += 1;}
else if (z>one) {
if (q1==0xfffffffe) q+=1;
q1+=2;
} else
q1 += (q1&1);
}
}
ix0 = (q>>1)+0x3fe00000;
ix1 = q1>>>1; // unsigned shift
if ((q&1)==1) ix1 |= sign;
ix0 += (m <<20);
// __HI(z) = ix0;
z = __HI(z, ix0);
// __LO(z) = ix1;
z = __LO(z, ix1);
return z;
}
}
/**
* cbrt(x)
* Return cube root of x

@ -0,0 +1,130 @@
/*
* Copyright (c) 2003, 2023, Oracle and/or its affiliates. All rights reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
* or visit www.oracle.com if you need additional information or have any
* questions.
*/
/*
* @test
* @bug 8302040
* @key randomness
* @library /test/lib
* @build jdk.test.lib.RandomFactory
* @build Tests
* @build FdlibmTranslit
* @build SqrtTests
* @run main SqrtTests
* @summary Tests for StrictMath.sqrt
*/
import jdk.test.lib.RandomFactory;
/**
* The tests in ../Math/SqrtTests.java test properties that should
* hold for any sqrt implementation, including the FDLIBM-based one
* required for StrictMath.sqrt. Therefore, the test cases in
* ../Math/SqrtTests.java are run against both the Math and
* StrictMath versions of sqrt. The role of this test is to verify
* that the FDLIBM sqrt algorithm is being used by running golden
* file tests on values that may vary from one conforming sqrt
* implementation to another.
*/
public class SqrtTests {
private SqrtTests(){}
public static void main(String... args) {
int failures = 0;
failures += testAgainstTranslit();
if (failures > 0) {
System.err.println("Testing sqrt incurred "
+ failures + " failures.");
throw new RuntimeException();
}
}
// Initialize shared random number generator
private static java.util.Random random = RandomFactory.getRandom();
/**
* Test StrictMath.sqrt against transliteration port of sqrt.
*/
private static int testAgainstTranslit() {
int failures = 0;
double x;
// Test just above subnormal threshold...
x = Double.MIN_NORMAL;
failures += testRange(x, Math.ulp(x), 1000);
// ... and just below subnormal threshold ...
x = Math.nextDown(Double.MIN_NORMAL);
failures += testRange(x, -Math.ulp(x), 1000);
// ... and near 1.0 ...
failures += testRangeMidpoint(1.0, Math.ulp(x), 2000);
// (Note: probes every-other value less than 1.0 due to
// change in the size of an ulp at 1.0.
// Probe near decision points in the FDLIBM algorithm.
double[] decisionPoints = {
Double.MIN_VALUE,
Double.MAX_VALUE,
};
for (double testPoint : decisionPoints) {
failures += testRangeMidpoint(testPoint, Math.ulp(testPoint), 1000);
}
x = Tests.createRandomDouble(random);
// Make the increment twice the ulp value in case the random
// value is near an exponent threshold. Don't worry about test
// elements overflowing to infinity if the starting value is
// near Double.MAX_VALUE.
failures += testRange(x, 2.0 * Math.ulp(x), 1000);
return failures;
}
private static int testRange(double start, double increment, int count) {
int failures = 0;
double x = start;
for (int i = 0; i < count; i++, x += increment) {
failures += testSqrtCase(x, FdlibmTranslit.sqrt(x));
}
return failures;
}
private static int testRangeMidpoint(double midpoint, double increment, int count) {
int failures = 0;
double x = midpoint - increment*(count / 2) ;
for (int i = 0; i < count; i++, x += increment) {
failures += testSqrtCase(x, FdlibmTranslit.sqrt(x));
}
return failures;
}
private static int testSqrtCase(double input, double expected) {
return Tests.test("StrictMath.sqrt(double)", input,
StrictMath::sqrt, expected);
}
}