Hyman Rosen <hymie@prolifics.com> wrote in message
news:t7g0vb2kg7.fsf@calumny.jyacc.com...
> According to 14.6.3, non-dependent names are bound at the point of
> use, so test would be looked up at this point, and not found. This
> code should not compile.

What is meant by point of use? If that mean when I first use the template,
then my call to test is defined, due to the using directives, and the code
should print B. At least that is how I would interpret the word use.

pth

namespace A { void test () { cout << "A" << endl; } }

namespace B { void test () { cout << "B" << endl; }}

template <class T> class sideffect {
public:  void tryit () { test (); }
};

using namespace A;
// using namespace B;

int main () {
  using B::test;

  sideffect<int> a;
  a.tryit ();

  return 0;
}





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"Patrick Hurley" <spam@hurleyhome.com> writes:
> What is meant by point of use? If that mean when I first use the template,
> then my call to test is defined, due to the using directives, and the code
> should print B. At least that is how I would interpret the word use.

It's the point where the template definition appears.
Look at the example in 14.6.3.

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Ok maybe I should buy the standard, but would you please help me with this
piece of code? I would have expected it to fail, as test does not exist in
the default namespace. If it should work, I would expect it to pick up
locallly defined use of B. In my available compilers this is not the case. I
get the A::test usage which was globally defined.

tia pth

#include <iostream>

using std::cout;
using std::endl;

namespace A {
  void test () {
    cout << "A" << endl;
  }
}

namespace B {
  void test () {
    cout << "B" << endl;
  }
}

template <class T>
class sideffect {
public:
  void tryit () {
    test ();
  }
private:
};

using namespace A;
// using namespace B;

int main () {
  using B::test;

  sideffect<int> a;
  a.tryit ();

  return 0;
}



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