Add num.parse.

sdk/lib/core/num.dart

 // Copyright (c) 2012, the Dart project authors.  Please see the AUTHORS file
 // for details. All rights reserved. Use of this source code is governed by a
 // BSD-style license that can be found in the LICENSE file.
 
 part of dart.core;
 
 /**
  * An integer or floating-point number.
  *
  * It is a compile-time error for any type other than [int] or [double]
@@ skipping 341 matching lines @@
    *     12.345e67.toString(); // 1.2345e+68
    *
    * Note: the conversion may round the output if the returned string
    * is accurate enough to uniquely identify the input-number.
    * For example the most precise representation of the [double] `9e59` equals
    * `"899999999999999918767229449717619953810131273674690656206848"`, but
    * this method returns the shorter (but still uniquely identifying) `"9e59"`.
    *
    */
   String toString();
+
+  /**
+   * Parses a string containing a number literal into a number.
+   *
+   * The input string should be either one accepted by [int.parse] (with no

[floitsch, 2013/11/25 12:50:53]

won't work in dartdoc: []()

Not sure if much better, but here is my attempt.

The method first tries to read the [input] as integer (similar to [int.parse]
without radix). If that fails, it tries to parse the [input] as double (as in
[double.parse]). If that fails, too, it invokes [onError] with [input].

If no [onError] is supplied, it defaults to a function that throws a
[FormatException].

[Lasse Reichstein Nielsen, 2013/11/25 13:05:16]

That is better. Done.

+   * radix) or by [double.parse].
+   * If it would be accepted by [int.parse],
+   * the result is the integer that `int.parse` would return.
+   * Otherwise the result is the same result that `double.parse` would return.
+   *
+   * If the input is acceptable by neither `int.parse` nor `double.parse`,
+   * the result is found by calling [onError] with the input.
+   * If no [onError] is suppled, a [FormatException] is thrown.

[floitsch, 2013/11/25 12:50:53]

supplied

[Lasse Reichstein Nielsen, 2013/11/25 13:05:16]

Replaced by your wording above.

+   *
+   * For any number `n`, this function satisfies `n == num.parse(n.toString())`.

[floitsch, 2013/11/25 12:50:53]

any number `n`, not NaN, ...

or use identical(n, num.parse(n.toString()))

[Lasse Reichstein Nielsen, 2013/11/25 13:05:16]

Used identical. That's what the test tests too.

+   */
+  static num parse(String input, [num onError(String input)]) {
+    String source = input.trim();
+    // TODO(lrn): Optimize to detect format and result type in one check.
+    num result = int.parse(source, onError: _returnNull);
+    if (result != null) return result;
+    result = double.parse(source, _returnNull);
+    if (result != null) return result;
+    if (onError == null) throw new FormatException(input);
+    return onError(input);
+  }
+
+  /** Helper function for [parse]. */
+  static _returnNull(_) => null;
 }