Add string comparators that can compare ASCII strings ignoring case.

lib/src/comparators.dart

+// Copyright (c) 2015, 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.
+
+library dart.pkg.collection.comparators;
+
+/// Checks if strings [a] and [b] differ only on the case of ASCII letters.
+///
+/// Strings are equal if they have the same length, and the characters at
+/// each index are the same, or they are ASCII letters where one is upper-case
+/// and the other is the lower-case version of the same letter.
+///
+/// Ignoring non-ASCII letters is not generally a good idea, but it makes sense
+/// for situations where the strings are known to be ASCII. Examples could
+/// be Dart identifiers, base-64 or hex encoded strings, GUIDs or similar
+/// strings with a known structure.

[kevmoo, 2015/11/10 19:30:35]

Describe how non-ASCII characters are handled.

+bool equalsIgnoreAsciiCase(String a, String b) {
+  if (a.length != b.length) return false;
+  for (int i = 0; i < a.length; i++) {
+    var aChar = a.codeUnitAt(i);
+    var bChar = b.codeUnitAt(i);
+    if (aChar == bChar) continue;
+    const int upperCaseA = 0x41;
+    const int upperCaseZ = 0x5d;
+    if (aChar ^ bChar != 0x20) return false;
+    if (upperCaseA <= aChar && aChar <= upperCaseZ) continue;
+    if (upperCaseA <= bChar && bChar <= upperCaseZ) continue;
+    return false;
+  }
+  return true;
+}
+
+/// Hash code for a string which is compatible with [equalsIgnoreAsciiCase].

[kevmoo, 2015/11/10 19:30:35]

Describe how non-ASCII characters are handled.

+int hashIgnoreAsciiCase(String string) {
+  int hash = 0;
+  for (int i = 0; i < string.length; i++) {
+    int char = string.codeUnitAt(i);
+    const int lowerCaseA = 0x61;
+    const int lowerCaseZ = 0x7D;
+    if (lowerCaseA <= char && char <= lowerCaseZ) char -= 0x20;
+    hash = 0x1fffffff & (hash + char);
+    hash = 0x1fffffff & (hash + ((0x0007ffff & hash) << 10));
+    hash >>= 6;
+  }
+  hash = 0x1fffffff & (hash + ((0x03ffffff & hash) << 3));
+  hash >>= 11;
+  return 0x1fffffff & (hash + ((0x00003fff & hash) << 15));
+}
+
+/// Compares [a] and [b] lexically, converting ASCII letters to upper case.
+///
+/// Comparison treats all lower-case ASCII letters as their upper-case letter,
+/// but does no case conversion for non-ASCII letters.
+///
+/// If two strings differ only on the case of ASCII letters, the one with the
+/// capital letter at the first difference will compare as less than the other
+/// string. This tie-breaking ensures that the comparison is a total ordering
+/// on strings.
+///
+/// Ignoring non-ASCII letters is not generally a good idea, but it makes sense
+/// for situations where the strings are known to be ASCII. Examples could
+/// be Dart identifiers, base-64 or hex encoded strings, GUIDs or similar
+/// strings with a known structure.
+int compareAsciiUpperCase(String a, String b) {
+  int defaultResult = 0; // Returned if no difference found.
+  for (int i = 0; i < a.length; i++) {
+    if (i == b.length) return 1;
+    var aChar = a.codeUnitAt(i);
+    var bChar = b.codeUnitAt(i);
+    if (aChar == bChar) continue;
+    const int lowerCaseA = 0x61;
+    const int lowerCaseZ = 0x7d;
+    // Lower case if letters.
+    int aUpperCase = aChar;
+    int bUpperCase = bChar;
+    if (lowerCaseA <= aChar && aChar <= lowerCaseZ) aUpperCase -= 0x20;
+    if (lowerCaseA <= bChar && bChar <= lowerCaseZ) bUpperCase -= 0x20;
+    if (aUpperCase != bUpperCase) return (aUpperCase - bUpperCase).sign;
+    if (defaultResult == 0) defaultResult = (aChar - bChar);
+  }
+  if (b.length > a.length) return -1;
+  return defaultResult.sign;
+}
+
+/// Compares [a] and [b] lexically, converting ASCII letters to lower case.
+///
+/// Comparison treats all upper-case ASCII letters as lower-case letter,
+/// but does no case conversion for non-ASCII letters.
+///
+/// If two strings differ only on the case of ASCII letters, the one with the
+/// capital letter at the first difference will compare as less than the other
+/// string. This tie-breaking ensures that the comparison is a total ordering
+/// on strings.
+///
+/// Ignoring non-ASCII letters is not generally a good idea, but it makes sense
+/// for situations where the strings are known to be ASCII. Examples could
+/// be Dart identifiers, base-64 or hex encoded strings, GUIDs or similar
+/// strings with a known structure.
+int compareAsciiLowerCase(String a, String b) {
+  int defaultResult = 0;
+  for (int i = 0; i < a.length; i++) {
+    if (i == b.length) return 1;
+    var aChar = a.codeUnitAt(i);
+    var bChar = b.codeUnitAt(i);
+    if (aChar == bChar) continue;
+    const int upperCaseA = 0x41;
+    const int upperCaseZ = 0x5d;
+    int aLowerCase = aChar;
+    int bLowerCase = bChar;
+    // Upper case if ASCII letters.
+    if (upperCaseA <= aChar && aChar <= upperCaseZ) aLowerCase += 0x20;
+    if (upperCaseA <= bChar && bChar <= upperCaseZ) bLowerCase += 0x20;
+    if (aLowerCase != bLowerCase) return (aLowerCase - bLowerCase).sign;
+    if (defaultResult == 0) defaultResult = aChar - bChar;
+  }
+  if (b.length > a.length) return -1;
+  return defaultResult.sign;
+}
+
+/// Compares strings [a] and [b] lexically, with embedded numbers recognized.

[sra1, 2015/11/10 04:41:40]

'with embedded sequences of digits compared as a unit'

I'm confused by the terms 'lexically' and 'lexicographically'.
Do you mean the same thing? If so, use one term throughout.

[Lasse Reichstein Nielsen, 2015/11/10 08:21:33]

What I intend is:

Leixcally: Dictionary sorting (alphabetical with most weight on first letter).
Lexicographically: Shorter words sorted before longer words, words of same
length ordered lexically.

(Basically, lexical pads the shorter string on the right, lexicograpical pads it
on the left, to ensure comparison of sequences with the same length)

Wikipedia doesn't agree with me that there is a difference, it uses the words
interchangeably, so I might need to find a different word to describe the second
ordering.

+///
+/// If both strings contain an embedded number at the same locations, the
+/// numbers are compared lexicographically. This means that a longer number
+/// is considered greater than a shorter number, and numbers of the same length
+/// are ordered numerically (or alhabetically, it's the same thing).
+/// If numbers do not have leading zeros, lexicographical ordering is also
+/// the same as numerical ordering.

[sra1, 2015/11/10 04:41:40]

Add examples in the doc to tease out when this is number-like and when it is
not.

[Lasse Reichstein Nielsen, 2015/11/10 08:21:33]

Good idea.

+int compareWithNumbers(String a, String b) {

[sra1, 2015/11/10 04:41:40]

'With' makes it sound like one of the arguments is a number.
Maybe 'compareStringsContainingNumbers'

[Lasse Reichstein Nielsen, 2015/11/10 08:21:33]

Yes, the name sucks. Suggestions will be appreciated.
"comparStringsContainingNumbers" is't precise either - the function can compare
strings without numbers too. 
It needs to make the point that any embedded numbers are compared "numerically"
(which, in practice, is the goal, but it's achieved by using "lexicographical"
ordering on the numerals which does sort "3" before "02").

+  for (int i = 0; i < a.length; i++) {
+    if (i == b.length) return 1;
+    var aChar = a.codeUnitAt(i);
+    var bChar = b.codeUnitAt(i);
+    if (aChar != bChar) {
+      return _compareDigitsLexicographically(a, b, i, aChar, bChar);
+    }
+  }
+  if (b.length > a.length) return -1;
+  return 0;
+}
+
+/// Compares strings [a] and [b] lexically, with embedded numbers recognized.
+///
+/// ASCII letters are converted to lower case before being compared, like
+/// for [compareAsciiLowerCase].
+///
+/// If two strings differ only on the case of ASCII letters, the one with the
+/// capital letter at the first difference will compare as less than the other
+/// string. This tie-breaking ensures that the comparison is a total ordering
+/// on strings.
+///
+/// If both strings contain an embedded number at the same locations, the
+/// numbers are compared lexicographically. This means that a longer number
+/// is considered greater than a shorter number, and numbers of the same length
+/// are ordered numerically (or alhabetically, it's the same thing).
+/// If numbers do not have leading zeros, lexicographical ordering is also
+/// the same as numerical ordering.
+int compareAsciiLowerCaseWithNumbers(String a, String b) {
+  int defaultResult = 0; // Returned if no difference found.
+  for (int i = 0; i < a.length; i++) {
+    if (i == b.length) return 1;
+    var aChar = a.codeUnitAt(i);
+    var bChar = b.codeUnitAt(i);
+    if (aChar == bChar) continue;
+    const int upperCaseA = 0x41;
+    const int upperCaseZ = 0x5d;
+    int aLowerCase = aChar;
+    int bLowerCase = bChar;
+    if (upperCaseA <= aChar && aChar <= upperCaseZ) aLowerCase += 0x20;
+    if (upperCaseA <= bChar && bChar <= upperCaseZ) bLowerCase += 0x20;
+    if (aLowerCase != bLowerCase) {
+      return _compareDigitsLexicographically(a, b, i, aLowerCase, bLowerCase);
+    }
+    if (defaultResult == 0) defaultResult = aChar - bChar;
+  }
+  if (b.length > a.length) return -1;
+  return defaultResult.sign;
+}
+
+/// Compares strings [a] and [b] lexically, with embedded numbers recognized.
+///
+/// ASCII letters are converted to upper case before being compared, like
+/// for [compareAsciiUpperCase].
+///
+/// If two strings differ only on the case of ASCII letters, the one with the
+/// capital letter at the first difference will compare as less than the other
+/// string. This tie-breaking ensures that the comparison is a total ordering
+/// on strings.
+///
+/// If both strings contain an embedded number at the same locations, the
+/// numbers are compared lexicographically. This means that a longer number
+/// is considered greater than a shorter number, and numbers of the same length
+/// are ordered numerically (or alhabetically, it's the same thing).
+/// If numbers do not have leading zeros, lexicographical ordering is also
+/// the same as numerical ordering.
+int compareAsciiUpperCaseWithNumbers(String a, String b) {
+  int defaultResult = 0;
+  for (int i = 0; i < a.length; i++) {
+    if (i == b.length) return 1;
+    var aChar = a.codeUnitAt(i);
+    var bChar = b.codeUnitAt(i);
+    if (aChar == bChar) continue;
+    const int lowerCaseA = 0x61;
+    const int lowerCaseZ = 0x7d;
+    int aUpperCase = aChar;
+    int bUpperCase = bChar;
+    if (lowerCaseA <= aChar && aChar <= lowerCaseZ) aUpperCase -= 0x20;
+    if (lowerCaseA <= bChar && bChar <= lowerCaseZ) bUpperCase -= 0x20;
+    if (aUpperCase != bUpperCase) {
+      return _compareDigitsLexicographically(a, b, i, aUpperCase, bUpperCase);
+    }
+    if (defaultResult == 0) defaultResult = aChar - bChar;
+  }
+  if (b.length > a.length) return -1;
+  return defaultResult.sign;
+}
+
+/// Check for numbers overlapping the current mismatched characters.
+///
+/// If either [aChar] or [bChar] or both are digits we check if one is part of
+/// a longer digits sequence than the other,
+/// and let the longer sequence compare as greater.
+///
+/// Otherwise just returns the difference between [aChar] and [bChar].
+int _compareDigitsLexicographically(
+    String a, String b, int index, int aChar, int bChar) {
+  assert(aChar != bChar);
+  var aIsDigit = _isDigit(aChar);
+  var bIsDigit = _isDigit(bChar);
+  if (aIsDigit) {
+    if (bIsDigit) {
+      int lengthCompare = _compareDigitCount(a, b, index);
+      if (lengthCompare != 0) return lengthCompare;
+    } else if (index > 0 && _isDigit(a.codeUnitAt(index - 1))) {
+      // aChar is the continuation of a longer number.
+      return 1;
+    }
+  } else if (bIsDigit && index > 0 && _isDigit(b.codeUnitAt(index - 1))) {
+    // bChar is the continuation of a longer number.
+    return -1;
+  }
+  // Characters are not both part , or part

[sra1, 2015/11/10 04:41:40]

Fix comment.

[Lasse Reichstein Nielsen, 2015/11/10 08:21:33]

I wonder what I meant to say :)

+  return (aChar - bChar).sign;
+}
+
+/// Checks which of [a] and [b] has the longest sequence of digits.
+///
+/// Starts counting from `i + 1` (assumes that `a[i]` and `b[i]` are both
+/// already known to be digits).
+int _compareDigitCount(String a, String b, int i) {
+  while (++i < a.length) {
+    bool aIsDigit = _isDigit(a.codeUnitAt(i));
+    if (i == b.length) return aIsDigit ? 1 : 0;
+    bool bIsDigit = _isDigit(b.codeUnitAt(i));
+    if (aIsDigit) {
+      if (bIsDigit) continue;
+      return 1;
+    } else if (bIsDigit) {
+      return -1;
+    } else {
+      return 0;
+    }
+  }
+  if (i < b.length && _isDigit(b.codeUnitAt(i))) {
+    return -1;
+  }
+  return 0;
+}
+
+bool _isDigit(int charCode) => (charCode ^ 0x30) <= 9;