1.1 --- /dev/null Thu Jan 01 00:00:00 1970 +0000
1.2 +++ b/js/src/tests/ecma/Math/15.8.2.18.js Wed Dec 31 06:09:35 2014 +0100
1.3 @@ -0,0 +1,131 @@
1.4 +/* -*- Mode: C++; tab-width: 2; indent-tabs-mode: nil; c-basic-offset: 2 -*- */
1.5 +/* This Source Code Form is subject to the terms of the Mozilla Public
1.6 + * License, v. 2.0. If a copy of the MPL was not distributed with this
1.7 + * file, You can obtain one at http://mozilla.org/MPL/2.0/. */
1.8 +
1.9 +
1.10 +/**
1.11 + File Name: 15.8.2.18.js
1.12 + ECMA Section: 15.8.2.18 tan( x )
1.13 + Description: return an approximation to the tan of the
1.14 + argument. argument is expressed in radians
1.15 + special cases:
1.16 + - if x is NaN result is NaN
1.17 + - if x is 0 result is 0
1.18 + - if x is -0 result is -0
1.19 + - if x is Infinity or -Infinity result is NaN
1.20 + Author: christine@netscape.com
1.21 + Date: 7 july 1997
1.22 +*/
1.23 +
1.24 +var SECTION = "15.8.2.18";
1.25 +var VERSION = "ECMA_1";
1.26 +startTest();
1.27 +var TITLE = "Math.tan(x)";
1.28 +var EXCLUDE = "true";
1.29 +
1.30 +writeHeaderToLog( SECTION + " "+ TITLE);
1.31 +
1.32 +new TestCase( SECTION,
1.33 + "Math.tan.length",
1.34 + 1,
1.35 + Math.tan.length );
1.36 +
1.37 +new TestCase( SECTION,
1.38 + "Math.tan()",
1.39 + Number.NaN,
1.40 + Math.tan() );
1.41 +
1.42 +new TestCase( SECTION,
1.43 + "Math.tan(void 0)",
1.44 + Number.NaN,
1.45 + Math.tan(void 0));
1.46 +
1.47 +new TestCase( SECTION,
1.48 + "Math.tan(null)",
1.49 + 0,
1.50 + Math.tan(null) );
1.51 +
1.52 +new TestCase( SECTION,
1.53 + "Math.tan(false)",
1.54 + 0,
1.55 + Math.tan(false) );
1.56 +
1.57 +new TestCase( SECTION,
1.58 + "Math.tan(NaN)",
1.59 + Number.NaN,
1.60 + Math.tan(Number.NaN) );
1.61 +
1.62 +new TestCase( SECTION,
1.63 + "Math.tan(0)",
1.64 + 0,
1.65 + Math.tan(0));
1.66 +
1.67 +new TestCase( SECTION,
1.68 + "Math.tan(-0)",
1.69 + -0,
1.70 + Math.tan(-0));
1.71 +
1.72 +new TestCase( SECTION,
1.73 + "Math.tan(Infinity)",
1.74 + Number.NaN,
1.75 + Math.tan(Number.POSITIVE_INFINITY));
1.76 +
1.77 +new TestCase( SECTION,
1.78 + "Math.tan(-Infinity)",
1.79 + Number.NaN,
1.80 + Math.tan(Number.NEGATIVE_INFINITY));
1.81 +
1.82 +new TestCase( SECTION,
1.83 + "Math.tan(Math.PI/4)",
1.84 + 1,
1.85 + Math.tan(Math.PI/4));
1.86 +
1.87 +new TestCase( SECTION,
1.88 + "Math.tan(3*Math.PI/4)",
1.89 + -1,
1.90 + Math.tan(3*Math.PI/4));
1.91 +
1.92 +new TestCase( SECTION,
1.93 + "Math.tan(Math.PI)",
1.94 + -0,
1.95 + Math.tan(Math.PI));
1.96 +
1.97 +new TestCase( SECTION,
1.98 + "Math.tan(5*Math.PI/4)",
1.99 + 1,
1.100 + Math.tan(5*Math.PI/4));
1.101 +
1.102 +new TestCase( SECTION,
1.103 + "Math.tan(7*Math.PI/4)",
1.104 + -1,
1.105 + Math.tan(7*Math.PI/4));
1.106 +
1.107 +new TestCase( SECTION,
1.108 + "Infinity/Math.tan(-0)",
1.109 + -Infinity,
1.110 + Infinity/Math.tan(-0) );
1.111 +
1.112 +/*
1.113 + Arctan (x) ~ PI/2 - 1/x for large x. For x = 1.6x10^16, 1/x is about the last binary digit of double precision PI/2.
1.114 + That is to say, perturbing PI/2 by this much is about the smallest rounding error possible.
1.115 +
1.116 + This suggests that the answer Christine is getting and a real Infinity are "adjacent" results from the tangent function. I
1.117 + suspect that tan (PI/2 + one ulp) is a negative result about the same size as tan (PI/2) and that this pair are the closest
1.118 + results to infinity that the algorithm can deliver.
1.119 +
1.120 + In any case, my call is that the answer we're seeing is "right". I suggest the test pass on any result this size or larger.
1.121 + = C =
1.122 +*/
1.123 +
1.124 +new TestCase( SECTION,
1.125 + "Math.tan(3*Math.PI/2) >= 5443000000000000",
1.126 + true,
1.127 + Math.tan(3*Math.PI/2) >= 5443000000000000 );
1.128 +
1.129 +new TestCase( SECTION,
1.130 + "Math.tan(Math.PI/2) >= 5443000000000000",
1.131 + true,
1.132 + Math.tan(Math.PI/2) >= 5443000000000000 );
1.133 +
1.134 +test();
