两个条件不能以相同的方式工作,我无法理解为什么

Question

我编写了一个函数来搜索数组中的char,如果找到则返回其后继,否则返回-1.然后,如果单词以元音结尾,则算法会添加辅音,反之亦然.

这段代码运行良好,即使文件的最后一个字:

changedChar = cipherChar(character, consonants, tConsonants);
if (changedChar != -1) charType = 'c';
else {
    changedChar = cipherChar(character, CONSONANTS, tConsonants);
    if (changedChar != -1) charType = 'c';
    else {
        changedChar = cipherChar(character, vowels, tVowels);
        if (changedChar != -1) charType = 'v';
        else {
            changedChar = cipherChar(character, VOWELS, tVowels);
            if (changedChar != -1) charType = 'v';
            else {
                changedChar = cipherChar(character, others, tOthers);
                if (changedChar != -1) charType = 'o';
                else {
                    changedChar = changeDigit(character);
                    if (changedChar != -1) charType = 'o';
                    else {
                        changedChar = cipherChar(character, punctuation, tPunctuation);
                        if (changedChar != -1) charType = 'o';
                    }
                }
            }
        }
    }
}
if (changedChar != -1) outFile << changedChar;
    if (searchChar(inFile.peek(), punctuation, tPunctuation) > -1)
        if (charType == 'v') {
            outFile << consonants[nVowel];
            nVowel < 4 ? nVowel++ : nVowel = 0;
        }
        else if (charType == 'c') {
            outFile << vowels[nConsonant];
    nConsonant < 20 ? nConsonant++ : nConsonant = 0;
        }

但是这个其他的不会在文件的最后一个字后添加一个额外的字母:

charType = 'c';
changedChar = cipherChar(character, consonants, tConsonants);
if (changedChar == -1) {
    changedChar = cipherChar(character, CONSONANTS, tConsonants);
    if (changedChar == -1) {
        charType = 'v';
        changedChar = cipherChar(character, vowels, tVowels);
        if (changedChar == -1) {
            changedChar = cipherChar(character, VOWELS, tVowels);
            if (changedChar == -1) {
                charType = 'o';
                changedChar = cipherChar(character, others, tOthers);
                if (changedChar == -1) {
                    changedChar = changeDigit(character);
                    if (changedChar == -1) changedChar = cipherChar(character, punctuation, tPunctuation);
                }
            }
        }
    }
}
if (changedChar != -1) outFile << changedChar;
    if (searchChar(inFile.peek(), punctuation, tPunctuation) > -1)
        if (charType == 'v') {
            outFile << consonants[nVowel];
            nVowel < 4 ? nVowel++ : nVowel = 0;
        }
        else if (charType == 'c') {
            outFile << vowels[nConsonant];
    nConsonant < 20 ? nConsonant++ : nConsonant = 0;
        }

为什么？我真的很困惑.

Answer 1

如果在逻辑中存在这么多条件,那么功能设计中应该存在一个缺陷.正确理解您的要求并尝试简化逻辑.