js/src/tests/ecma/Math/15.8.2.18.js@6474c204b198
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.
1 /* -*- Mode: C++; tab-width: 2; indent-tabs-mode: nil; c-basic-offset: 2 -*- */
2 /* This Source Code Form is subject to the terms of the Mozilla Public
3 * License, v. 2.0. If a copy of the MPL was not distributed with this
4 * file, You can obtain one at http://mozilla.org/MPL/2.0/. */
7 /**
8 File Name: 15.8.2.18.js
9 ECMA Section: 15.8.2.18 tan( x )
10 Description: return an approximation to the tan of the
11 argument. argument is expressed in radians
12 special cases:
13 - if x is NaN result is NaN
14 - if x is 0 result is 0
15 - if x is -0 result is -0
16 - if x is Infinity or -Infinity result is NaN
17 Author: christine@netscape.com
18 Date: 7 july 1997
19 */
21 var SECTION = "15.8.2.18";
22 var VERSION = "ECMA_1";
23 startTest();
24 var TITLE = "Math.tan(x)";
25 var EXCLUDE = "true";
27 writeHeaderToLog( SECTION + " "+ TITLE);
29 new TestCase( SECTION,
30 "Math.tan.length",
31 1,
32 Math.tan.length );
34 new TestCase( SECTION,
35 "Math.tan()",
36 Number.NaN,
37 Math.tan() );
39 new TestCase( SECTION,
40 "Math.tan(void 0)",
41 Number.NaN,
42 Math.tan(void 0));
44 new TestCase( SECTION,
45 "Math.tan(null)",
46 0,
47 Math.tan(null) );
49 new TestCase( SECTION,
50 "Math.tan(false)",
51 0,
52 Math.tan(false) );
54 new TestCase( SECTION,
55 "Math.tan(NaN)",
56 Number.NaN,
57 Math.tan(Number.NaN) );
59 new TestCase( SECTION,
60 "Math.tan(0)",
61 0,
62 Math.tan(0));
64 new TestCase( SECTION,
65 "Math.tan(-0)",
66 -0,
67 Math.tan(-0));
69 new TestCase( SECTION,
70 "Math.tan(Infinity)",
71 Number.NaN,
72 Math.tan(Number.POSITIVE_INFINITY));
74 new TestCase( SECTION,
75 "Math.tan(-Infinity)",
76 Number.NaN,
77 Math.tan(Number.NEGATIVE_INFINITY));
79 new TestCase( SECTION,
80 "Math.tan(Math.PI/4)",
81 1,
82 Math.tan(Math.PI/4));
84 new TestCase( SECTION,
85 "Math.tan(3*Math.PI/4)",
86 -1,
87 Math.tan(3*Math.PI/4));
89 new TestCase( SECTION,
90 "Math.tan(Math.PI)",
91 -0,
92 Math.tan(Math.PI));
94 new TestCase( SECTION,
95 "Math.tan(5*Math.PI/4)",
96 1,
97 Math.tan(5*Math.PI/4));
99 new TestCase( SECTION,
100 "Math.tan(7*Math.PI/4)",
101 -1,
102 Math.tan(7*Math.PI/4));
104 new TestCase( SECTION,
105 "Infinity/Math.tan(-0)",
106 -Infinity,
107 Infinity/Math.tan(-0) );
109 /*
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.
111 That is to say, perturbing PI/2 by this much is about the smallest rounding error possible.
113 This suggests that the answer Christine is getting and a real Infinity are "adjacent" results from the tangent function. I
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
115 results to infinity that the algorithm can deliver.
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.
118 = C =
119 */
121 new TestCase( SECTION,
122 "Math.tan(3*Math.PI/2) >= 5443000000000000",
123 true,
124 Math.tan(3*Math.PI/2) >= 5443000000000000 );
126 new TestCase( SECTION,
127 "Math.tan(Math.PI/2) >= 5443000000000000",
128 true,
129 Math.tan(Math.PI/2) >= 5443000000000000 );
131 test();
