k_tanf.rs - source
Description: Source of the Rust file `/Users/erlendbasso/.cargo/registry/src/github.com-1ecc6299db9ec823/libm-0.2.6/src/math/k_tanf.rs`.
Keywords: rust, rustlang, rust-lang

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    /* origin: FreeBSD /usr/src/lib/msun/src/k_tan.c */
    /*
     * ====================================================
     * Copyright 2004 Sun Microsystems, Inc.  All Rights Reserved.
     *
     * Permission to use, copy, modify, and distribute this
     * software is freely granted, provided that this notice
     * is preserved.
     * ====================================================
     */

    /* |tan(x)/x - t(x)| < 2**-25.5 (~[-2e-08, 2e-08]). */
    const T: [f64; 6] = [
        0.333331395030791399758,   /* 0x15554d3418c99f.0p-54 */
        0.133392002712976742718,   /* 0x1112fd38999f72.0p-55 */
        0.0533812378445670393523,  /* 0x1b54c91d865afe.0p-57 */
        0.0245283181166547278873,  /* 0x191df3908c33ce.0p-58 */
        0.00297435743359967304927, /* 0x185dadfcecf44e.0p-61 */
        0.00946564784943673166728, /* 0x1362b9bf971bcd.0p-59 */
    ];

    #[cfg_attr(all(test, assert_no_panic), no_panic::no_panic)]
    pub(crate) fn k_tanf(x: f64, odd: bool) -> f32 {
        let z = x * x;
        /*
         * Split up the polynomial into small independent terms to give
         * opportunities for parallel evaluation.  The chosen splitting is
         * micro-optimized for Athlons (XP, X64).  It costs 2 multiplications
         * relative to Horner's method on sequential machines.
         *
         * We add the small terms from lowest degree up for efficiency on
         * non-sequential machines (the lowest degree terms tend to be ready
         * earlier).  Apart from this, we don't care about order of
         * operations, and don't need to to care since we have precision to
         * spare.  However, the chosen splitting is good for accuracy too,
         * and would give results as accurate as Horner's method if the
         * small terms were added from highest degree down.
         */
        let mut r = T[4] + z * T[5];
        let t = T[2] + z * T[3];
        let w = z * z;
        let s = z * x;
        let u = T[0] + z * T[1];
        r = (x + s * u) + (s * w) * (t + w * r);
        (if odd { -1. / r } else { r }) as f32
    }