comparing words to those in a vector?

[sofa]

hey everyone, this is my first post, and I have been learning c++ for a few months now (and boy, it's confusing!)

Right now, I"m trying to select words. If they are in a vector of words that are 'taboo' they are rejected. if not, they are placed into a map.

So far, I have been looping through the vector as i compare to each element, but for some reason, this is screwing up the count. Is comparing to each element the only way to go?
instead of counting each 'good' word, it prints out the map, but every word with an associative number incrementing by 4 is displayed.

[Danish]

I don't understand. You're comparing words (strings?) to words in a vector, then moving them to a map(?) once you're done. That sounds like a logic error. Could we see the problematic code?

[Valmont]

Declaring a string vector is simple:

vector <string>vecStringTabooWords;

The size of a vector is determined by: vecStringTabooWords.size();

So iterating a vector goes like:

for(int i = 0; i < vecStringTabooWords.size(); i++)

Comparing goes like: vecStringTabooWords.[i] ==some_string

Well, I don't know what else to type. Just post the code and the error messages if any.

[Unicorn]

To look for something in a container, you can use the 'find' function from the <algorithm> header. To look for a word in your vector, you can use it like this:

vector<string>::iterator found (std::find (taboo.begin(), taboo.end(), some_word));

'found' will now point to the word you were looking for, or to 'taboo.end()' if it wasn't found.

If all the words in your taboo list are unique, consider using a set instead. A set will sort its elements for you, and you can be sure you are not wasting precious memory by storing the same word twice. Also, because the container is sorted, looking for a certain word is a lot faster.

set<string> taboo_set;
taboo_set.insert ("foo");

if (taboo_set.count ("bar") == 0)
  // The word "bar" is not in the map