uClibc now has a math library. muahahahaha!

 -Erik

1 files changed, 122 insertions, 0 deletions

diff --git a/libm/double/mod2pi.c b/libm/double/mod2pi.c
new file mode 100644
index 000000000..057954a9b
--- /dev/null
+++ b/libm/double/mod2pi.c
@@ -0,0 +1,122 @@
+/* Program to test range reduction of trigonometry functions
+ *
+ * -- Steve Moshier
+ */
+
+#include <math.h>
+#ifdef ANSIPROT
+extern double floor ( double );
+extern double ldexp ( double, int );
+extern double sin ( double );
+#else
+double floor(), ldexp(), sin();
+#endif
+
+#define TPI 6.283185307179586476925
+
+main()
+{
+char s[40];
+double a, n, t, x, y, z;
+int lflg;
+
+x = TPI/4.0;
+t = 1.0;
+
+loop:
+
+t = 2.0 * t;
+
+/* Stop testing at a point beyond which the integer part of
+ * x/2pi cannot be represented exactly by a double precision number.
+ * The library trigonometry functions will probably give up long before
+ * this point is reached.
+ */
+if( t > 1.0e16 )
+	exit(0);
+
+/* Adjust the following to choose a nontrivial x
+ * where test function(x) has a slope of about 1 or more.
+ */
+x = TPI * t  + 0.5;
+
+z = x;
+lflg = 0;
+
+inlup:
+
+/* floor() returns the largest integer less than its argument.
+ * If you do not have this, or AINT(), then you may convert x/TPI
+ * to a long integer and then back to double; but in that case
+ * x will be limited to the largest value that will fit into a
+ * long integer.
+ */
+n = floor( z/TPI );
+
+/* Carefully subtract 2 pi n from x.
+ * This is done by subtracting n * 2**k in such a way that there
+ * is no arithmetic cancellation error at any step.  The k are the
+ * bits in the number 2 pi.
+ *
+ * If you do not have ldexp(), then you may multiply or
+ * divide n by an appropriate power of 2 after each step.
+ * For example:
+ *  a = z - 4*n;
+ *  a -= 2*n;
+ *  n /= 4;
+ *  a -= n;   n/4
+ *  n /= 8;
+ *  a -= n;   n/32
+ * etc.
+ * This will only work if division by a power of 2 is exact.
+ */
+
+a = z - ldexp(n, 2);	/* 4n */
+a -= ldexp( n, 1);	/* 2n */
+a -= ldexp( n, -2 );	/* n/4 */
+a -= ldexp( n, -5 );	/* n/32 */
+a -= ldexp( n, -9 );	/* n/512 */
+a += ldexp( n, -15 );	/* add n/32768 */
+a -= ldexp( n, -17 );	/* n/131072 */
+a -= ldexp( n, -18 );
+a -= ldexp( n, -20 );
+a -= ldexp( n, -22 );
+a -= ldexp( n, -24 );
+a -= ldexp( n, -28 );
+a -= ldexp( n, -32 );
+a -= ldexp( n, -37 );
+a -= ldexp( n, -39 );
+a -= ldexp( n, -40 );
+a -= ldexp( n, -42 );
+a -= ldexp( n, -46 );
+a -= ldexp( n, -47 );
+
+/* Subtract what is left of 2 pi n after all the above reductions.
+ */
+a -= 2.44929359829470635445e-16 * n;
+
+/* If the test is extended too far, it is possible
+ * to have chosen the wrong value of n.  The following
+ * will fix that, but at some reduction in accuracy.
+ */
+if( (a > TPI) || (a < -1e-11) )
+	{
+	z = a;
+	lflg += 1;
+	printf( "Warning! Reduction failed on first try.\n" );
+	goto inlup;
+	}
+if( a < 0.0 )
+	{
+	printf( "Warning! Reduced value < 0\n" );
+	a += TPI;
+	}
+
+/* Compute the test function at x and at a = x mod 2 pi.
+ */
+y = sin(x);
+z = sin(a);
+printf( "sin(%.15e) error = %.3e\n", x, y-z );
+goto loop;
+}
+