Question

I tried the code from this question C++ std::transform() and toupper() ..why does this fail?

#include <iostream>
#include <algorithm>

int main() {
  std::string s="hello";
  std::string out;
  std::transform(s.begin(), s.end(), std::back_inserter(out), std::toupper);
  std::cout << "hello in upper case: " << out << std::endl;
}

Theoretically it should’ve worked as it’s one of the examples in Josuttis’ book, but it doesn’t compile http://ideone.com/aYnfv.

Why did GCC complain:

no matching function for call to ‘transform(
    __gnu_cxx::__normal_iterator<char*, std::basic_string
        <char, std::char_traits<char>, std::allocator<char> > >, 
    __gnu_cxx::__normal_iterator<char*, std::basic_string
        <char, std::char_traits<char>, std::allocator<char> > >, 
    std::back_insert_iterator<std::basic_string
        <char, std::char_traits<char>, std::allocator<char> > >,
    <unresolved overloaded function type>)’

Am I missing something here? Is it GCC related problem?

Answer 1

Just use ::toupper instead of std::toupper. That is, toupper defined in the global namespace, instead of the one defined in std namespace.

std::transform(s.begin(), s.end(), std::back_inserter(out), ::toupper);

Its working : http://ideone.com/XURh7

Reason why your code is not working : there is another overloaded function toupper in the namespace std which is causing problem when resolving the name, because compiler is unable to decide which overload you’re referring to, when you simply pass std::toupper. That is why the compiler is saying unresolved overloaded function type in the error message, which indicates the presence of overload(s).

So to help the compiler in resolving to the correct overload, you’ve to cast std::toupper as

(int (*)(int))std::toupper

That is, the following would work:

//see the last argument, how it is casted to appropriate type
std::transform(s.begin(), s.end(), std::back_inserter(out),(int (*)(int))std::toupper);

Check it out yourself: http://ideone.com/8A6iV