Finishing Up
What about vowels? Rule #3 also states that otherwise-duplicate encodings separated by a vowel (not h or w) get coded twice.
| c2/39/SoundexTest.cpp | |
| | TEST_F(SoundexEncoding, DoesNotCombineDuplicateEncodingsSeparatedByVowels) { |
| | ASSERT_THAT(soundex.encode("Jbob"), Eq("J110")); |
| | } |
Once again we solve our challenge by declaring what we want to accomplish. We change the conditional expression in encodeLetter to append a digit only if it is not a duplicate or if the last letter is a vowel. This declaration drives a few corresponding changes.
| c2/39/Soundex.h | |
| | void encodeTail(std::string& encoding, const std::string& word) const { |
* | for (auto i = 1u; i < word.length(); i++) |
* | if (!isComplete(encoding)) |
* | encodeLetter(encoding, word[i], word[i - 1]); |
| | } |
| | |
* | void encodeLetter(std::string& encoding, char letter, char lastLetter) const { |
| | auto digit = encodedDigit(letter); |
| | if (digit != NotADigit && |
* | (digit != lastDigit(encoding) || isVowel(lastLetter))) |
| | encoding += digit; |
| | } |
| | |
* | bool isVowel(char letter) const { |
* | return |
* | std::string("aeiouy").find(lower(letter)) != std::string::npos; |
* | } |
Is passing in the previous letter the best way to do this? It’s direct and expressive. We’ll stick with it for now.
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