【问题标题】:Declaring variadic templates of pointer to member functions声明指向成员函数的指针的可变参数模板
【发布时间】:2015-09-17 14:02:55
【问题描述】:

我想构建自己的单元测试库,我想在其中设置测试用例,如下所示:

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的指针的可变参数模板:

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 及以上版本出现以下错误:

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代码!

声明成员函数指针的可变参数模板的正确方法是什么?

Here is the full code

【问题讨论】:

    标签: c++ unit-testing c++11 g++ variadic-templates


    【解决方案1】:

    变化:

    template <typename Test, void(Test::*First)(), void(Test::*...Rest)()>
    class test_case <Test, &Test::First, &Test::...Rest>
    

    进入:

    template <typename Test, void(Test::*First)(), void(Test::*...Rest)()>
    class test_case <Test, First, Rest...>
    

    还有:

    test_case<Test, &Test::...Rest>()(args...);
    

    进入:

    test_case<Test, Rest...>()(args...);
    

    【讨论】:

    • Piotr,为什么 g++-4.7 不给&amp;Test::...Rest 语法错误?以及它如何能够构建实际上是语法错误的东西?
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