将函数模板特化传递给可变参数模板函数

Question

I have no problem passing the address of a function template specialization to a regular template function:

template <typename T>
void f(T) {}

template <typename A, typename B>
void foo(A, B) {}

int main()
{
    foo(&f<int>, &f<float>);
}

However, when I try to pass the same specializations to a variadic template:

template <typename T>
void f(T) {}

template <typename... A>
void bar(A...) {}

int main()
{
    bar(&f<int>, &f<float>);
}

I get the following compiler errors with GCC (I tried 4.6.1 and 4.7.0):

test.cpp: In function 'int main()':
test.cpp:9:27: error: no matching function for call to 'bar(<unresolved overloaded function type>, <unresolved overloaded function type>)'
test.cpp:9:27: note: candidate is:
test.cpp:5:6: note: template<class ... A> void bar(A ...)
test.cpp:5:6: note:   template argument deduction/substitution failed:

Why am I getting these errors?

Answer 1

看起来它可能是GCC中的一个可能在GCC 4.6.2中修复的错误 （我说可能是因为它不完全相同，但确实涉及获取可变参数模板函数的地址）。