声明成员函数指针的可变模板

我想建立我自己的单元测试库，我想设置测试用例如下:

template <typename... Args>
std::string concatenate(Args&&... args);
class my_test : public unit_test::test {
public:
    my_test(int i, float f, double d) : i_(i), f_(f), d_(d) { }
    void test1() { assert_true(i_ % 5 == 0, concatenate("i(", i_, ") not divisible by 5")); }
    void test2() { assert_true(i_ > 0, concatenate("i(", i_, ") not greater than 0")); }
    void test3() { assert_true(i_ % 2 == 0, concatenate("i(", i_, ") not divisible by 2")); }
private:
    int i_;
    float f_;
    double d_;
};
int main()
{
    unit_test::test_case<my_test,
        &my_test::test1
        &my_test::test2
        &my_test::test3> my_test_case;
    result r = my_test_case(1, 1.0f, 1.0);
}

为了能够定义test_case模板类，我需要能够声明指向成员函数指针的可变模板:

class result {
    unsigned int num_failures_;
    unsigned int num_tests_;
};
template <typename Test, void(Test::*...MemFns)()>
class test_case;

不幸的是，g++-4.8及以上版本给出以下错误:

main.cpp:137:52: error: template argument 3 is invalid
 class test_case <Test, &Test::First, &Test::...Rest> {
                                                    ^
main.cpp: In function 'int main(int, char**)':
main.cpp:194:28: error: template argument 2 is invalid
             &my_test::test3>()(1, 1.0f, 1.0);

令人惊讶的是，g++-4.7编译并运行了一些无效的代码!

声明成员函数指针的可变模板的正确方法是什么?

完整代码