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Issue 520693004: Dart version of double.parse that works for numbers that can be represented (Closed) Base URL: https://dart.googlecode.com/svn/branches/bleeding_edge/dart
Patch Set: Remove unused variabl. Created 6 years, 2 months ago
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1 // Copyright (c) 2012, the Dart project authors. Please see the AUTHORS file 1 // Copyright (c) 2012, the Dart project authors. Please see the AUTHORS file
2 // for details. All rights reserved. Use of this source code is governed by a 2 // for details. All rights reserved. Use of this source code is governed by a
3 // BSD-style license that can be found in the LICENSE file. 3 // BSD-style license that can be found in the LICENSE file.
4 // Dart core library. 4 // Dart core library.
5 5
6 // VM implementation of double. 6 // VM implementation of double.
7 7
8 patch class double { 8 patch class double {
9 9
10 static double _nativeParse(String str, 10 static double _nativeParse(String str,
11 int start, int end) native "Double_parse"; 11 int start, int end) native "Double_parse";
12 12
13 static double _tryParseDouble(var str, var start, var end) {
14 assert(start < end);
15 const int _DOT = 0x2e; // '.'
16 const int _ZERO = 0x30; // '0'
17 const int _MINUS = 0x2d; // '-'
18 const int _N = 0x4e; // 'N'
19 const int _a = 0x61; // 'a'
20 const int _I = 0x49; // 'I'
21 const int _e = 0x65; // 'e'
22 int exponent = 0;
23 // Set to non-zero if a digit is seen. Avoids accepting ".".
24 int digitsSeen = 0;
Vyacheslav Egorov (Google) 2014/10/02 14:27:41 please change this to boolean
Lasse Reichstein Nielsen 2014/10/02 14:36:44 Done. Please optimize booleans :)
25 // Added to exponent for each digit. Set to -1 when seeing '.'.
26 int exponentDelta = 0;
27 double doubleValue = 0.0;
28 double sign = 1.0;
29 int firstChar = str.codeUnitAt(start);
30 if (firstChar == _MINUS) {
31 sign = -1.0;
32 start++;
33 if (start == end) return null;
34 firstChar = str.codeUnitAt(start);
35 }
36 if (firstChar == _I) {
37 if (end == start + 8 && str.startsWith("nfinity", start + 1)) {
38 return sign * double.INFINITY;
39 }
40 return null;
41 }
42 if (firstChar == _N) {
43 if (end == start + 3 && str.codeUnitAt(start + 1) == _a
Vyacheslav Egorov (Google) 2014/10/02 14:27:40 1 line && 1 line && 1 line
Lasse Reichstein Nielsen 2014/10/02 14:36:44 Done.
44 && str.codeUnitAt(start + 2) == _N) {
45 return double.NAN;
46 }
47 return null;
48 }
49
50 int firstDigit = firstChar ^ _ZERO;
51 if (firstDigit <= 9) {
52 start++;
53 doubleValue = firstDigit.toDouble();
54 digitsSeen = 1;
55 }
56 for (int i = start; i < end; i++) {
57 int c = str.codeUnitAt(i);
58 int digit = c ^ _ZERO; // '0'-'9' characters are now 0-9 integers.
59 if (digit <= 9) {
60 doubleValue = doubleValue * 10 + digit;
61 if (doubleValue >= 9007199254740992.0) return null; // Maybe unprecise.
Vyacheslav Egorov (Google) 2014/10/02 14:27:41 I would make this a const with a good name.
Lasse Reichstein Nielsen 2014/10/02 14:36:44 Done.
62 exponent += exponentDelta;
63 digitsSeen = 1;
64 } else if ((c == _DOT) && (exponentDelta == 0)) {
Vyacheslav Egorov (Google) 2014/10/02 14:27:40 there is an inconsistency when you parenthesize ==
Lasse Reichstein Nielsen 2014/10/02 14:36:43 Done, without the parentheses.
65 exponentDelta = -1;
66 } else if ((c | 0x20) == _e) {
67 i++;
68 if (i == end) return null;
69 int expPart = int._tryParseSmi(str, i, end - 1);
Vyacheslav Egorov (Google) 2014/10/02 14:27:41 I find this end - 1 confusing. Is it because our p
Lasse Reichstein Nielsen 2014/10/02 14:36:44 The _tryParseSmi is unconventional in having end i
70 if (expPart == null) return null;
71 exponent += expPart;
72 break;
73 } else {
74 return null;
75 }
76 }
77 if (digitsSeen == 0) return null; // No digits.
78 if (exponent == 0) return sign * doubleValue;
79 // Powers of 10 up to 10^22 are exact doubles.
Vyacheslav Egorov (Google) 2014/10/02 14:27:41 It would be good to use something more elaborate i
Lasse Reichstein Nielsen 2014/10/02 14:36:44 Added comment.
80 const P10 = const [
81 1.0, /* 0 */
82 10.0,
83 100.0,
84 1000.0,
85 10000.0,
86 100000.0, /* 5 */
87 1000000.0,
88 10000000.0,
89 100000000.0,
90 1000000000.0,
91 10000000000.0, /* 10 */
92 100000000000.0,
93 1000000000000.0,
94 10000000000000.0,
95 100000000000000.0,
96 1000000000000000.0, /* 15 */
97 10000000000000000.0,
98 100000000000000000.0,
99 1000000000000000000.0,
100 10000000000000000000.0,
101 100000000000000000000.0, /* 20 */
102 1000000000000000000000.0,
103 10000000000000000000000.0,
104 ];
105 if (exponent < 0) {
106 if (exponent < -22) return null;
Vyacheslav Egorov (Google) 2014/10/02 14:27:41 maybe use P10.length here instead of 22?
Lasse Reichstein Nielsen 2014/10/02 14:36:44 That does appear to be just as fast. Done.
107 return sign * (doubleValue / P10[-exponent]);
108 }
109 if (exponent > 22) return null;
Vyacheslav Egorov (Google) 2014/10/02 14:27:41 Ditto
Lasse Reichstein Nielsen 2014/10/02 14:36:44 Done.
110 return sign * (doubleValue * P10[exponent]);
111 }
112
13 static double _parse(var str) { 113 static double _parse(var str) {
14 int len = str.length; 114 int len = str.length;
15 int start = str._firstNonWhitespace(); 115 int start = str._firstNonWhitespace();
16 if (start == len) return null; // All whitespace. 116 if (start == len) return null; // All whitespace.
17 int end = str._lastNonWhitespace() + 1; 117 int end = str._lastNonWhitespace() + 1;
18 assert(start < end); 118 assert(start < end);
19 119 var result = _tryParseDouble(str, start, end);
120 if (result != null) return result;
20 return _nativeParse(str, start, end); 121 return _nativeParse(str, start, end);
21 } 122 }
22 123
23 /* patch */ static double parse(String str, 124 /* patch */ static double parse(String str,
24 [double onError(String str)]) { 125 [double onError(String str)]) {
25 var result = _parse(str); 126 var result = _parse(str);
26 if (result == null) { 127 if (result == null) {
27 if (onError == null) throw new FormatException("Invalid double", str); 128 if (onError == null) throw new FormatException("Invalid double", str);
28 return onError(str); 129 return onError(str);
29 } 130 }
30 return result; 131 return result;
31 } 132 }
32 } 133 }
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