JavaScriptCore-7601.1.46.3.tar.gz

[apple/javascriptcore.git] / tests / mozilla / ecma / Math / 15.8.2.18.js

/* The contents of this file are subject to the Netscape Public
 * License Version 1.1 (the "License"); you may not use this file
 * except in compliance with the License. You may obtain a copy of
 * the License at http://www.mozilla.org/NPL/
 *
 * Software distributed under the License is distributed on an "AS
 * IS" basis, WITHOUT WARRANTY OF ANY KIND, either express or
 * implied. See the License for the specific language governing
 * rights and limitations under the License.
 *
 * The Original Code is Mozilla Communicator client code, released March
 * 31, 1998.
 *
 * The Initial Developer of the Original Code is Netscape Communications
 * Corporation. Portions created by Netscape are
 * Copyright (C) 1998 Netscape Communications Corporation. All
 * Rights Reserved.
 *
 * Contributor(s): 
 * 
 */
/**
    File Name:          15.8.2.18.js
    ECMA Section:       15.8.2.18 tan( x )
    Description:        return an approximation to the tan of the
                        argument.  argument is expressed in radians
                        special cases:
                        - if x is NaN           result is NaN
                        - if x is 0             result is 0
                        - if x is -0            result is -0
                        - if x is Infinity or -Infinity result is NaN
    Author:             christine@netscape.com
    Date:               7 july 1997
*/

    var SECTION = "15.8.2.18";
    var VERSION = "ECMA_1";
    startTest();
    var TITLE   = "Math.tan(x)";
    var EXCLUDE = "true";

    writeHeaderToLog( SECTION + " "+ TITLE);

    var testcases = getTestCases();
    test();

function getTestCases() {
    var array = new Array();
    var item = 0;

    array[item++] = new TestCase( SECTION,  "Math.tan.length",          1,              Math.tan.length );

    array[item++] = new TestCase( SECTION,  "Math.tan()",               Number.NaN,      Math.tan() );
    array[item++] = new TestCase( SECTION,  "Math.tan(void 0)",         Number.NaN,     Math.tan(void 0));
    array[item++] = new TestCase( SECTION,  "Math.tan(null)",           0,              Math.tan(null) );
    array[item++] = new TestCase( SECTION,  "Math.tan(false)",          0,              Math.tan(false) );

    array[item++] = new TestCase( SECTION,  "Math.tan(NaN)",            Number.NaN,     Math.tan(Number.NaN) );
    array[item++] = new TestCase( SECTION,  "Math.tan(0)",              0,	            Math.tan(0));
    array[item++] = new TestCase( SECTION,  "Math.tan(-0)",             -0,         	Math.tan(-0));
    array[item++] = new TestCase( SECTION,  "Math.tan(Infinity)",       Number.NaN,     Math.tan(Number.POSITIVE_INFINITY));
    array[item++] = new TestCase( SECTION,  "Math.tan(-Infinity)",      Number.NaN,     Math.tan(Number.NEGATIVE_INFINITY));
    array[item++] = new TestCase( SECTION,  "Math.tan(Math.PI/4)",      1,              Math.tan(Math.PI/4));
    array[item++] = new TestCase( SECTION,  "Math.tan(3*Math.PI/4)",    -1,             Math.tan(3*Math.PI/4));
    array[item++] = new TestCase( SECTION,  "Math.tan(Math.PI)",        -0,             Math.tan(Math.PI));
    array[item++] = new TestCase( SECTION,  "Math.tan(5*Math.PI/4)",    1,              Math.tan(5*Math.PI/4));
    array[item++] = new TestCase( SECTION,  "Math.tan(7*Math.PI/4)",    -1,             Math.tan(7*Math.PI/4));
    array[item++] = new TestCase( SECTION,  "Infinity/Math.tan(-0)",    -Infinity,      Infinity/Math.tan(-0) );

/*
    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.
    That is to say, perturbing PI/2 by this much is about the smallest rounding error possible.

    This suggests that the answer Christine is getting and a real Infinity are "adjacent" results from the tangent function.  I
    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
    results to infinity that the algorithm can deliver.

    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.
    = C =
*/
    array[item++] = new TestCase( SECTION,  "Math.tan(3*Math.PI/2) >= 5443000000000000",   true,   Math.tan(3*Math.PI/2) >= 5443000000000000 );
    array[item++] = new TestCase( SECTION,  "Math.tan(Math.PI/2) >= 5443000000000000",      true,   Math.tan(Math.PI/2) >= 5443000000000000 );

    return ( array );
}
function test() {
    for ( tc=0; tc < testcases.length; tc++ ) {
        testcases[tc].passed = writeTestCaseResult(
                            testcases[tc].expect,
                            testcases[tc].actual,
                            testcases[tc].description +" = "+
                            testcases[tc].actual );

        testcases[tc].reason += ( testcases[tc].passed ) ? "" : "wrong value ";
    }
    stopTest();
    return ( testcases );
}