debugging c++ program with <set> containers - c++

Question

The problem is to find all vowels all even words in a sentence share, in other words these vowels have to be met in any even word in the sentence.
But i when i enter for example: "ewedyua aiuye dswidje ieuayj eeee eeeui dajhdfjcne aodijsbfe".
I get: e i u
but "e i" is expected as the last even word doesn't contain 'u' (I use " " just for separation here in the text and don't use them in the output)
The program:
int main(){
string str;
char ch = ' ';
set<char> strSet;
set<char> resultSet;
set<char> tempSet;
int count = 1;
int i = 0;
cout << "Enter a line: ";
getline(cin, str);
str = delOverSpace(str); // delete excessive gaps<br>
do {
ch = str.at(i);
if(((count % 2) == 0) && (ch != ' ')){ // this is an even word and not a gap
if(isVowel(upperToLower(ch))) // this is a vowel
tempSet.insert(upperToLower(ch));
}
if (ch == ' ') { // if we've passed through the word add inforamtion on it
if(((count % 2) == 0) && (count / 2) == 1)
strSet.insert(tempSet.begin(), tempSet.end());
else if (((count % 2) == 0) && (count / 2) != 1){
set_intersection(
strSet.begin(),strSet.end(), tempSet.begin(), tempSet.end(),
insert_iterator<set<char> >(resultSet, resultSet.begin())
);
strSet.clear();
tempSet.clear();
strSet.insert(resultSet.begin(), resultSet.end());
resultSet.clear();
}
count++;
}
i++;
}while(ch != '.');
if (count == 2) cout << "Only one word was entered" << endl;
else if (strSet.empty()) cout << "No vowels were found" << endl;
else {
copy(strSet.begin(), strSet.end(), ostream_iterator<char>(cout, " "));
cout << endl;
}
return 0;
}

Answer 1

Try not to use one counter variable for multiple purposes.
For example you could use count for the current character count within the current word and words for word count.
In your example, you're making count work too hard.
Decide what you are counting. Do you want to count every character?
Or do you want to count the characters in the current word?
When incrementing a counter, try to put this next to the logic that needs it.
Eg, i++ should go right after you've used it to get the next character.
And count should be incremented in the block that handles letters (ie not space, not dot).
If the counters get too far away from what they are keeping track of, you can introduce bugs.
Move the space check up first and also check for end of sentence with .
Clear indentation helps to follow the code - I use 4 characters, but 2 or 3 is ok.
1 character indents are hard to read.
Use blank lines to separate different chunks of logic.
You might want to add logic to handle punctuation like ,, ?, ! etc.
When you're processing an input string which is separated into words, it's often easier to handle the end of input condition outside the loop.
In this case, the loop exits after it picks up the last letter.
If that is not a . then you'll have an unprocessed word sitting in tempSet.
You can try and anticipate this within the loop, but it seems more natural to exit the loop and check if there is anything left to process.
If you separate your logic into clearly separated blocks, you can set breakpoints inside each block.
You'll then know what situation you're handling. You can break inside end-of-word logic and check tempSet and count.
Are they what you expect?
And in the else part you can check ch going in and verify that it is classified as a vowel properly.
int main() {
string str;
char ch = ' ';
set<char> strSet;
set<char> tempSet;
int count = 0, words = 0, i = 0;
cout << "Enter a line: ";
getline(cin, str);
str = delOverSpace(str); // delete excessive gaps<br>
while ( i < str.size() && ch != '.' ) {
// Get next character
ch = str.at(i);
i++;
if (ch == ' ' || ch == '.' ) { // if we've passed through the word add inforamtion on it
if( count > 0 && ((count % 2) == 0)) {
strSet.insert(tempSet.begin(), tempSet.end());
words++;
}
count = 0; // reset the char counter
tempSet.clear(); // clear the temp vowel set
}
else {
// letter - not a space or a dot, increase the character count
count++;
if(isVowel(upperToLower(ch))) // this is a vowel
tempSet.insert(upperToLower(ch));
}
}
// pick up any unprocessed word if we hit end of input without seeing '.'
if (count > 0 && ((count % 2) == 0)) {
strSet.insert(tempSet.begin(), tempSet.end());
words++;
}
if (words == 0)
cout << "No words entered" << endl;
else if (words == 1)
cout << "Only one word was entered" << endl;
else if (strSet.empty())
cout << "No vowels were found" << endl;
else {
copy(strSet.begin(), strSet.end(), ostream_iterator<char>(cout, " "));
cout << endl;
}
return 0;
}