Visual C++ Compiler allows dependent-name as a type without “typename”?

问题

Today one of my friends told me that the following code compiles well on his Visual Studio 2008:

#include <vector>
struct A
{
  static int const const_iterator = 100;
};
int i;
template <typename T>
void PrintAll(const T & obj)
{
  T::const_iterator *i;
}
int main()
{
  std::vector<int> v;
  A a;
  PrintAll(a);
  PrintAll(v);
  return 0;
}

I usually use g++, and it always refuse to pass the second PrintAll() call. As I know, for this problem, g++ is doing the standard way translating a template.

So, is my knowledge wrong, or is it a extension of VS2008?

回答1:

This is not an extension at all.

VC++ never implemented the two phases interpretation properly:

At the point of definition, parse the template and determine all non-dependent names
At the point of instantiation, check that the template produces valid code

VC++ never implemented the first phase... it's inconvenient since it means not only that it accepts code that is non-compliant but also that it produces an altogether different code in some situations.

void foo(int) { std::cout << "int" << std::endl; }

template <class T> void tfoo() { foo(2.0); }

void foo(double) { std::cout << "double" << std::endl; }

int main(int argc, char* argv[])
{
  tfoo<Dummy>();
}

With this code:

compliant compilers will print "int", because it was the only definition available at the point of definition of the template and the resolution of foo does not depend on T.
VC++ will print "double", because it never bothered with phase 1

It might seem stupid as far as differences go, but if you think about the number of includes you have in a large program, there is a risk that someone will introduce an overload after your template code... and BAM :/

回答2:

I'm not sure "extension" is exactly how I'd describe VC++ in this respect, but yes, gcc has better conformance in this regard.

来源：https://stackoverflow.com/questions/2974780/visual-c-compiler-allows-dependent-name-as-a-type-without-typename

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