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Diffstat (limited to 'js/src/tests/ecma/String/15.5.4.7-1.js')
| -rw-r--r-- | js/src/tests/ecma/String/15.5.4.7-1.js | 185 |
1 files changed, 185 insertions, 0 deletions
diff --git a/js/src/tests/ecma/String/15.5.4.7-1.js b/js/src/tests/ecma/String/15.5.4.7-1.js new file mode 100644 index 000000000..8559bea69 --- /dev/null +++ b/js/src/tests/ecma/String/15.5.4.7-1.js @@ -0,0 +1,185 @@ +/* -*- tab-width: 2; indent-tabs-mode: nil; js-indent-level: 2 -*- */ +/* This Source Code Form is subject to the terms of the Mozilla Public + * License, v. 2.0. If a copy of the MPL was not distributed with this + * file, You can obtain one at http://mozilla.org/MPL/2.0/. */ + + +/** + File Name: 15.5.4.7-1.js + ECMA Section: 15.5.4.7 String.prototype.lastIndexOf( searchString, pos) + Description: + + If the given searchString appears as a substring of the result of + converting this object to a string, at one or more positions that are + at or to the left of the specified position, then the index of the + rightmost such position is returned; otherwise -1 is returned. If position + is undefined or not supplied, the length of this string value is assumed, + so as to search all of the string. + + When the lastIndexOf method is called with two arguments searchString and + position, the following steps are taken: + + 1.Call ToString, giving it the this value as its argument. + 2.Call ToString(searchString). + 3.Call ToNumber(position). (If position is undefined or not supplied, this step produces the value NaN). + 4.If Result(3) is NaN, use +; otherwise, call ToInteger(Result(3)). + 5.Compute the number of characters in Result(1). + 6.Compute min(max(Result(4), 0), Result(5)). + 7.Compute the number of characters in the string that is Result(2). + 8.Compute the largest possible integer k not larger than Result(6) such that k+Result(7) is not greater + than Result(5), and for all nonnegative integers j less than Result(7), the character at position k+j of + Result(1) is the same as the character at position j of Result(2); but if there is no such integer k, then + compute the value -1. + + 1.Return Result(8). + + Note that the lastIndexOf function is intentionally generic; it does not require that its this value be a + String object. Therefore it can be transferred to other kinds of objects for use as a method. + + Author: christine@netscape.com + Date: 2 october 1997 +*/ +var SECTION = "15.5.4.7-1"; +var VERSION = "ECMA_1"; +startTest(); +var TITLE = "String.protoype.lastIndexOf"; + +var TEST_STRING = new String( " !\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~" ); + +writeHeaderToLog( SECTION + " "+ TITLE); + +var j = 0; + +for ( k = 0, i = 0x0021; i < 0x007e; i++, j++, k++ ) { + new TestCase( SECTION, + "String.lastIndexOf(" +String.fromCharCode(i)+ ", 0)", + -1, + TEST_STRING.lastIndexOf( String.fromCharCode(i), 0 ) ); +} + +for ( k = 0, i = 0x0020; i < 0x007e; i++, j++, k++ ) { + new TestCase( SECTION, + "String.lastIndexOf("+String.fromCharCode(i)+ ", "+ k +")", + k, + TEST_STRING.lastIndexOf( String.fromCharCode(i), k ) ); +} + +for ( k = 0, i = 0x0020; i < 0x007e; i++, j++, k++ ) { + new TestCase( SECTION, + "String.lastIndexOf("+String.fromCharCode(i)+ ", "+k+1+")", + k, + TEST_STRING.lastIndexOf( String.fromCharCode(i), k+1 ) ); +} + +for ( k = 9, i = 0x0021; i < 0x007d; i++, j++, k++ ) { + new TestCase( SECTION, + + "String.lastIndexOf("+(String.fromCharCode(i) + + String.fromCharCode(i+1)+ + String.fromCharCode(i+2)) +", "+ 0 + ")", + LastIndexOf( TEST_STRING, String.fromCharCode(i) + + String.fromCharCode(i+1)+String.fromCharCode(i+2), 0), + TEST_STRING.lastIndexOf( (String.fromCharCode(i)+ + String.fromCharCode(i+1)+ + String.fromCharCode(i+2)), + 0 ) ); +} + +for ( k = 0, i = 0x0020; i < 0x007d; i++, j++, k++ ) { + new TestCase( SECTION, + "String.lastIndexOf("+(String.fromCharCode(i) + + String.fromCharCode(i+1)+ + String.fromCharCode(i+2)) +", "+ k +")", + k, + TEST_STRING.lastIndexOf( (String.fromCharCode(i)+ + String.fromCharCode(i+1)+ + String.fromCharCode(i+2)), + k ) ); +} +for ( k = 0, i = 0x0020; i < 0x007d; i++, j++, k++ ) { + new TestCase( SECTION, + "String.lastIndexOf("+(String.fromCharCode(i) + + String.fromCharCode(i+1)+ + String.fromCharCode(i+2)) +", "+ k+1 +")", + k, + TEST_STRING.lastIndexOf( (String.fromCharCode(i)+ + String.fromCharCode(i+1)+ + String.fromCharCode(i+2)), + k+1 ) ); +} +for ( k = 0, i = 0x0020; i < 0x007d; i++, j++, k++ ) { + new TestCase( SECTION, + "String.lastIndexOf("+ + (String.fromCharCode(i) + + String.fromCharCode(i+1)+ + String.fromCharCode(i+2)) +", "+ (k-1) +")", + LastIndexOf( TEST_STRING, String.fromCharCode(i) + + String.fromCharCode(i+1)+String.fromCharCode(i+2), k-1), + TEST_STRING.lastIndexOf( (String.fromCharCode(i)+ + String.fromCharCode(i+1)+ + String.fromCharCode(i+2)), + k-1 ) ); +} + +new TestCase( SECTION, "String.lastIndexOf(" +TEST_STRING + ", 0 )", 0, TEST_STRING.lastIndexOf( TEST_STRING, 0 ) ); + +// new TestCase( SECTION, "String.lastIndexOf(" +TEST_STRING + ", 1 )", 0, TEST_STRING.lastIndexOf( TEST_STRING, 1 )); + +new TestCase( SECTION, "String.lastIndexOf(" +TEST_STRING + ")", 0, TEST_STRING.lastIndexOf( TEST_STRING )); + +print( "TEST_STRING = new String(\" !\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~\")" ); + +test(); + +function LastIndexOf( string, search, position ) { + string = String( string ); + search = String( search ); + + position = Number( position ) + + if ( isNaN( position ) ) { + position = Infinity; + } else { + position = ToInteger( position ); + } + + result5= string.length; + result6 = Math.min(Math.max(position, 0), result5); + result7 = search.length; + + if (result7 == 0) { + return Math.min(position, result5); + } + + result8 = -1; + + for ( k = 0; k <= result6; k++ ) { + if ( k+ result7 > result5 ) { + break; + } + for ( j = 0; j < result7; j++ ) { + if ( string.charAt(k+j) != search.charAt(j) ){ + break; + } else { + if ( j == result7 -1 ) { + result8 = k; + } + } + } + } + + return result8; +} +function ToInteger( n ) { + n = Number( n ); + if ( isNaN(n) ) { + return 0; + } + if ( Math.abs(n) == 0 || Math.abs(n) == Infinity ) { + return n; + } + + var sign = ( n < 0 ) ? -1 : 1; + + return ( sign * Math.floor(Math.abs(n)) ); +} |