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

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);
like image 767
sivabudh Avatar asked Sep 28 '09 20:09

sivabudh


2 Answers

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');

like image 199
hrnt Avatar answered Oct 16 '22 23:10

hrnt


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.

like image 2
Michael Krelin - hacker Avatar answered Oct 16 '22 22:10

Michael Krelin - hacker