Dart version of double.parse that works for numbers that can be represented

runtime/lib/double_patch.dart

Index: runtime/lib/double_patch.dart
diff --git a/runtime/lib/double_patch.dart b/runtime/lib/double_patch.dart
index d8e0cbd0c8450825ce4460e4af4e0a60d7291337..5becf9a8ae17cc7ecc710d6caceaa2baa813072d 100644
--- a/runtime/lib/double_patch.dart
+++ b/runtime/lib/double_patch.dart
@@ -10,13 +10,114 @@ patch class double {
   static double _nativeParse(String str,
                              int start, int end) native "Double_parse";
 
+  static double _tryParseDouble(var str, var start, var end) {
+    assert(start < end);
+    const int _DOT = 0x2e;    // '.'
+    const int _ZERO = 0x30;   // '0'
+    const int _MINUS = 0x2d;  // '-'
+    const int _N = 0x4e;      // 'N'
+    const int _a = 0x61;      // 'a'
+    const int _I = 0x49;      // 'I'
+    const int _e = 0x65;      // 'e'
+    int exponent = 0;
+    // Set to non-zero if a digit is seen. Avoids accepting ".".
+    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 :)

+    // Added to exponent for each digit. Set to -1 when seeing '.'.
+    int exponentDelta = 0;
+    double doubleValue = 0.0;
+    double sign = 1.0;
+    int firstChar = str.codeUnitAt(start);
+    if (firstChar == _MINUS) {
+      sign = -1.0;
+      start++;
+      if (start == end) return null;
+      firstChar = str.codeUnitAt(start);
+    }
+    if (firstChar == _I) {
+      if (end == start + 8 && str.startsWith("nfinity", start + 1)) {
+        return sign * double.INFINITY;
+      }
+      return null;
+    }
+    if (firstChar == _N) {
+      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.

+                           && str.codeUnitAt(start + 2) == _N) {
+        return double.NAN;
+      }
+      return null;
+    }
+
+    int firstDigit = firstChar ^ _ZERO;
+    if (firstDigit <= 9) {
+      start++;
+      doubleValue = firstDigit.toDouble();
+      digitsSeen = 1;
+    }
+    for (int i = start; i < end; i++) {
+      int c = str.codeUnitAt(i);
+      int digit = c ^ _ZERO;  // '0'-'9' characters are now 0-9 integers.
+      if (digit <= 9) {
+        doubleValue = doubleValue * 10 + digit;
+        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.

+        exponent += exponentDelta;
+        digitsSeen = 1;
+      } else if ((c == _DOT) && (exponentDelta == 0)) {

[Vyacheslav Egorov (Google), 2014/10/02 14:27:40]

there is an inconsistency when you parenthesize ==-comparisons and when you
don't. please either always write (a == b) && (c == d) or a == b && c == d.  

[Lasse Reichstein Nielsen, 2014/10/02 14:36:43]

Done, without the parentheses.

+        exponentDelta = -1;
+      } else if ((c | 0x20) == _e) {
+        i++;
+        if (i == end) return null;
+        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 parsing functions being
inconsistent with end being inclusive/exclusive?

Can we settle on end being either inclusive or exclusive everywhere? At least
add a comment here.

[Lasse Reichstein Nielsen, 2014/10/02 14:36:44]

The _tryParseSmi is unconventional in having end inclusive (it made sense at the
time, because we calculate the "last non-whitespace index", so we can use that
as argument directly on the fast path. 

We should probably clean that up, now that I use it from elsewhere too.

+        if (expPart == null) return null;
+        exponent += expPart;
+        break;
+      } else {
+        return null;
+      }
+    }
+    if (digitsSeen == 0) return null;  // No digits.
+    if (exponent == 0) return sign * doubleValue;
+    // 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 instead of 'exact' it's a bit
confusing. 

[Lasse Reichstein Nielsen, 2014/10/02 14:36:44]

Added comment.

+    const P10 = const [
+                             1.0,  /*  0 */
+                            10.0,
+                           100.0,
+                          1000.0,
+                         10000.0,
+                        100000.0,  /*  5 */
+                       1000000.0,
+                      10000000.0,
+                     100000000.0,
+                    1000000000.0,
+                   10000000000.0,  /* 10 */
+                  100000000000.0,
+                 1000000000000.0,
+                10000000000000.0,
+               100000000000000.0,
+              1000000000000000.0,  /*  15 */
+             10000000000000000.0,
+            100000000000000000.0,
+           1000000000000000000.0,
+          10000000000000000000.0,
+         100000000000000000000.0,  /*  20 */
+        1000000000000000000000.0,
+       10000000000000000000000.0,
+    ];
+    if (exponent < 0) {
+      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.

+      return sign * (doubleValue / P10[-exponent]);
+    }
+    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.

+    return sign * (doubleValue * P10[exponent]);
+  }
+
   static double _parse(var str) {
     int len = str.length;
     int start = str._firstNonWhitespace();
     if (start == len) return null;  // All whitespace.
     int end = str._lastNonWhitespace() + 1;
     assert(start < end);
-
+    var result = _tryParseDouble(str, start, end);
+    if (result != null) return result;
     return _nativeParse(str, start, end);
   }