C++ std::transform() and toupper() ..why does this fail?

Question

I have 2 std::string. I just want to, given the input string:

1. capitalize every letter
2. assign the capitalized letter to the output string.

How come this works:

  std::string s="hello";
  std::string out;
  std::transform(s.begin(), s.end(), std::back_inserter(out), std::toupper);

but this doesn't (results in a program crash)?

  std::string s="hello";
  std::string out;
  std::transform(s.begin(), s.end(), out.begin(), std::toupper);

because this works (at least on the same string:

  std::string s="hello";
  std::string out;
  std::transform(s.begin(), s.end(), s.begin(), std::toupper);

Answer 1

There is no space in out . C++ algorithms do not grow their target containers automatically. You must either make the space yourself, or use a inserter adaptor.

To make space in out , do this:

out.resize(s.length());

[edit] Another option is to create the output string with correct size with this constructor.

std::string out(s.length(), 'X');

Answer 2

I'd say that the iterator returned by out.begin() is not valid after a couple of increments for the empty string. After the first ++ it's ==out.end() , then the behavior after the next increment is undefined.

After all this exactly what insert iterator is for.

Answer 3

Thats the sense of a backinserter: It inserts elements to a container. using begin(), you pass a iterator to a empty container and modify invalid iterators.

I am sorry - my edits interfered with your comments. I first posted something wrong accidentally.