前言就是上個 Task 囉，假設句子為
您終於回來了
您終於回來了
您終於回來了
我終於回來了
那麼程式原來的 output 是
您終於回來了,3
終於回來了,4
而原來的 bug 版本跑過之後，會將
終於回來了,4
吃掉，只剩第一個，因為做了修正。修改之後就好了 
原先的:
for (j = 1; j < wi; j++) {
// This function check "abcd,4,2" and "bcd,3,2"
// then set "bcd,3,2" to "bcd,3,0"
if (word_freq[j].wend == word_freq[j-1].wend) {
word_freq[j].freq = 0;
}
}
// This function check "abcd,4,2" and "bcd,3,2"
// then set "bcd,3,2" to "bcd,3,0"
if (word_freq[j].wend == word_freq[j-1].wend) {
word_freq[j].freq = 0;
}
}
改過之後的:
for (j = 1, i = j - 1; j < wi; j++) {
// This function check "abcd,4,2" and "bcd,3,2"
// then set "bcd,3,2" to "bcd,3,0"
if (word_freq[j].wend == word_freq[i].wend && word_freq[j].freq == word_freq[i].freq) {
word_freq[j].freq = 0;
} else
i = j; // j will become j+1 later (for loop)
}
// This function check "abcd,4,2" and "bcd,3,2"
// then set "bcd,3,2" to "bcd,3,0"
if (word_freq[j].wend == word_freq[i].wend && word_freq[j].freq == word_freq[i].freq) {
word_freq[j].freq = 0;
} else
i = j; // j will become j+1 later (for loop)
}
結論是，這種有大量 input/output 的程式還真難 debug，尤其在結果大部份都是合理的情況下，更是難以分辨。