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 method first tries to read the [input] as integer (similar to
+   * [int.parse] without a radix).
+   * If that fails, it tries to parse the [input] as a double (similar to
+   * [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].
+   *
+   * For any number `n`, this function satisfies
+   * `identical(n, num.parse(n.toString()))`.
+   */
+  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;
 }