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Is a static member function of a function-scoped struct of a function template considered a dependent scope?

Tags:

c++

MSVC 9 and g++-4.5 disagree over the use of typename within nested::baz below. Which is correct?

template<typename T>
  struct foo
{
  typedef T type;
};

template<typename T>
  typename foo<T>::type
    bar(T x)
{
  struct nested
  {
    inline static typename foo<T>::type baz(T x)
    {
      typename foo<T>::type result;
      return result;
    }
  };

  return nested::baz(x);
}

int main()
{
  int x;
  bar(x);

  return 0;
}

MSVC's output:

$ cl test.cpp
Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 15.00.30729.01 for 80x86
Copyright (C) Microsoft Corporation.  All rights reserved.

test.cpp
test.cpp(15) : error C2899: typename cannot be used outside a template declaration

g++-4.5 issues no error, but if I remove the offending typename, I receive an error message:

$ g++-4.5 test.cpp
test.cpp: In static member function 'static typename foo<T>::type bar(T)::nested::baz(T)':
test.cpp:15:7: error: need 'typename' before 'foo<T>::type' because 'foo<T>' is a dependent scope
test.cpp:15:20: error: expected ';' before 'result'
test.cpp:16:14: error: 'result' was not declared in this scope
test.cpp: In static member function 'static typename foo<T>::type bar(T)::nested::baz(T) [with T = int, typename foo<T>::type = int]':
test.cpp:20:23:   instantiated from 'typename foo<T>::type bar(T) [with T = int, typename foo<T>::type = int]'
test.cpp:26:8:   instantiated from here
test.cpp:15:7: error: dependent-name 'foo<T>::type' is parsed as a non-type, but instantiation yields a type
test.cpp:15:7: note: say 'typename foo<T>::type' if a type is meant

Which is correct in this instance?

like image 746
Jared Hoberock Avatar asked Nov 05 '22 09:11

Jared Hoberock


1 Answers

MSVC seems to be at fault; see this related question provided by jagansai above: typename outside of template


Here is a workaround for bar which both compilers are happy with:

template<typename T>
  typename foo<T>::type
    bar(T x)
{
  typedef typename foo<T>::type result_type;

  struct nested
  {
    inline static result_type baz(T x)
    {
      result_type result;
      return result;
    }
  };

  return nested::baz(x);
}
like image 123
Jared Hoberock Avatar answered Nov 09 '22 11:11

Jared Hoberock