js/src/tests/ecma/Math/15.8.2.18.js

Wed, 31 Dec 2014 06:09:35 +0100

author
Michael Schloh von Bennewitz <michael@schloh.com>
date
Wed, 31 Dec 2014 06:09:35 +0100
changeset 0
6474c204b198
permissions
-rw-r--r--

Cloned upstream origin tor-browser at tor-browser-31.3.0esr-4.5-1-build1
revision ID fc1c9ff7c1b2defdbc039f12214767608f46423f for hacking purpose.

michael@0 1 /* -*- Mode: C++; tab-width: 2; indent-tabs-mode: nil; c-basic-offset: 2 -*- */
michael@0 2 /* This Source Code Form is subject to the terms of the Mozilla Public
michael@0 3 * License, v. 2.0. If a copy of the MPL was not distributed with this
michael@0 4 * file, You can obtain one at http://mozilla.org/MPL/2.0/. */
michael@0 5
michael@0 6
michael@0 7 /**
michael@0 8 File Name: 15.8.2.18.js
michael@0 9 ECMA Section: 15.8.2.18 tan( x )
michael@0 10 Description: return an approximation to the tan of the
michael@0 11 argument. argument is expressed in radians
michael@0 12 special cases:
michael@0 13 - if x is NaN result is NaN
michael@0 14 - if x is 0 result is 0
michael@0 15 - if x is -0 result is -0
michael@0 16 - if x is Infinity or -Infinity result is NaN
michael@0 17 Author: christine@netscape.com
michael@0 18 Date: 7 july 1997
michael@0 19 */
michael@0 20
michael@0 21 var SECTION = "15.8.2.18";
michael@0 22 var VERSION = "ECMA_1";
michael@0 23 startTest();
michael@0 24 var TITLE = "Math.tan(x)";
michael@0 25 var EXCLUDE = "true";
michael@0 26
michael@0 27 writeHeaderToLog( SECTION + " "+ TITLE);
michael@0 28
michael@0 29 new TestCase( SECTION,
michael@0 30 "Math.tan.length",
michael@0 31 1,
michael@0 32 Math.tan.length );
michael@0 33
michael@0 34 new TestCase( SECTION,
michael@0 35 "Math.tan()",
michael@0 36 Number.NaN,
michael@0 37 Math.tan() );
michael@0 38
michael@0 39 new TestCase( SECTION,
michael@0 40 "Math.tan(void 0)",
michael@0 41 Number.NaN,
michael@0 42 Math.tan(void 0));
michael@0 43
michael@0 44 new TestCase( SECTION,
michael@0 45 "Math.tan(null)",
michael@0 46 0,
michael@0 47 Math.tan(null) );
michael@0 48
michael@0 49 new TestCase( SECTION,
michael@0 50 "Math.tan(false)",
michael@0 51 0,
michael@0 52 Math.tan(false) );
michael@0 53
michael@0 54 new TestCase( SECTION,
michael@0 55 "Math.tan(NaN)",
michael@0 56 Number.NaN,
michael@0 57 Math.tan(Number.NaN) );
michael@0 58
michael@0 59 new TestCase( SECTION,
michael@0 60 "Math.tan(0)",
michael@0 61 0,
michael@0 62 Math.tan(0));
michael@0 63
michael@0 64 new TestCase( SECTION,
michael@0 65 "Math.tan(-0)",
michael@0 66 -0,
michael@0 67 Math.tan(-0));
michael@0 68
michael@0 69 new TestCase( SECTION,
michael@0 70 "Math.tan(Infinity)",
michael@0 71 Number.NaN,
michael@0 72 Math.tan(Number.POSITIVE_INFINITY));
michael@0 73
michael@0 74 new TestCase( SECTION,
michael@0 75 "Math.tan(-Infinity)",
michael@0 76 Number.NaN,
michael@0 77 Math.tan(Number.NEGATIVE_INFINITY));
michael@0 78
michael@0 79 new TestCase( SECTION,
michael@0 80 "Math.tan(Math.PI/4)",
michael@0 81 1,
michael@0 82 Math.tan(Math.PI/4));
michael@0 83
michael@0 84 new TestCase( SECTION,
michael@0 85 "Math.tan(3*Math.PI/4)",
michael@0 86 -1,
michael@0 87 Math.tan(3*Math.PI/4));
michael@0 88
michael@0 89 new TestCase( SECTION,
michael@0 90 "Math.tan(Math.PI)",
michael@0 91 -0,
michael@0 92 Math.tan(Math.PI));
michael@0 93
michael@0 94 new TestCase( SECTION,
michael@0 95 "Math.tan(5*Math.PI/4)",
michael@0 96 1,
michael@0 97 Math.tan(5*Math.PI/4));
michael@0 98
michael@0 99 new TestCase( SECTION,
michael@0 100 "Math.tan(7*Math.PI/4)",
michael@0 101 -1,
michael@0 102 Math.tan(7*Math.PI/4));
michael@0 103
michael@0 104 new TestCase( SECTION,
michael@0 105 "Infinity/Math.tan(-0)",
michael@0 106 -Infinity,
michael@0 107 Infinity/Math.tan(-0) );
michael@0 108
michael@0 109 /*
michael@0 110 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.
michael@0 111 That is to say, perturbing PI/2 by this much is about the smallest rounding error possible.
michael@0 112
michael@0 113 This suggests that the answer Christine is getting and a real Infinity are "adjacent" results from the tangent function. I
michael@0 114 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
michael@0 115 results to infinity that the algorithm can deliver.
michael@0 116
michael@0 117 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.
michael@0 118 = C =
michael@0 119 */
michael@0 120
michael@0 121 new TestCase( SECTION,
michael@0 122 "Math.tan(3*Math.PI/2) >= 5443000000000000",
michael@0 123 true,
michael@0 124 Math.tan(3*Math.PI/2) >= 5443000000000000 );
michael@0 125
michael@0 126 new TestCase( SECTION,
michael@0 127 "Math.tan(Math.PI/2) >= 5443000000000000",
michael@0 128 true,
michael@0 129 Math.tan(Math.PI/2) >= 5443000000000000 );
michael@0 130
michael@0 131 test();

mercurial