1 files changed, 0 insertions, 122 deletions
diff --git a/libm/double/mod2pi.c b/libm/double/mod2pi.c
deleted file mode 100644
index 057954a9b..000000000
--- a/libm/double/mod2pi.c
+++ /dev/null
@@ -1,122 +0,0 @@
-/* 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;
-}
-