Search() algorythm

I'm trying to use the algorithm Search() with strings.
I get an error at the line where i marked.
Could somebody please give me a hint what am i doing wrong? Thanks in advance!

Code:

int main () {
  vector<string> myvector ;
  vector<int>::iterator it;
  
  string str = "search";
  string match = "ea";
  myvector.push_back(str);
  
  it = search (myvector.begin(), myvector.end(), match.begin(), match.end());   // error

  if (it != myvector.end())
    cout << "Match found at position: " << int(it - myvector.begin()) << endl;
  else
    cout << "Match not found." << endl;
  
  system("pause");
  return 0;
}

Sorry for the typo: algorithm.

It appears that you are trying to compare single characters (what string iterators point to) with whole strings (what vector<string> iterator points to), and eventually assign the result to a vector<int> iterator?!

If you just want to find a substring in a string, then string has the find method.

Thank you, i'm gonna check it if i can do it with find().

Just out of interest, what would you have expected the output of your program to be?

I would have expected something like: "Match found at position: 1" .

So if i understand you well we should always compare the same types in C++.
So maybe i should compare the first element of the vector to a string.
Like this
search (myvector[0].begin(), myvector[0].end(), match.begin(), match.end());

So, have you worked out a solution?

Like this
search (myvector[0].begin(), myvector[0].end(), match.begin(), match.end());

That looks more like it.

However, if you only want to find a string in another string there may be easier ways: string reference

Well for my final program i think im gonna use find() because it indeed does sound easier.

For this specific program i dont understand why cant i assign search() to the iterator::it,
when: "search() returns an iterator to the beginning of that matching subrange" .

Is it still an iterator of vector<int>? Since you search in a string, the result of search can only be assigned to string::iterator.

Hmm thanks, i thought an iterator is always an int because its a number.
I've changed the program but still get the error.

Code:

int main () {
  vector<string> myvector ;
  vector<string>::iterator it;
  
  string str = "search";
  string match = "ea";
  myvector.push_back(str);
  
  it = search (myvector[0].begin(), myvector[0].end(), match.begin(), match.end());   // error

  if (it != myvector[0].end())
    cout << "Match found at position: " << int(it - myvector[0].begin()) << endl;
  else
    cout << "Match not found." << endl;
  
  system("pause");
  return 0;
}

if you have an old college text book.

they might have a bubble sort sample

Originally Posted by Ducky

Hmm thanks, i thought an iterator is always an int because its a number.

An iterator is never just a number or an int. It is usually contains a pointer, and possibly has other members as well.

@t3chn0n3rd bubble sort sample That rings me a bell, gotta look it up.

@iMalc sounds like iterator is a complicated thing.

bubble sort sample That rings me a bell, gotta look it up.

Go ahead and look it up to refresh your knowledge, but in this case I do not think it applies. If it did, you should use something like std::sort() from <algorithm> instead of implementing bubble sort.

sounds like iterator is a complicated thing.

It can be complicated, or it can be as simple as a pointer, since pointers are iterators.

You are searching a string, not a vector. The string just happens to be put into a vector in this case (don't know why you need it there). Therefore the result needs to go to a string::iterator, not a vector<string>::iterator.