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c++ - 声明指向成员函数的可变参数模板

转载 作者:行者123 更新时间:2023-11-30 05:38:38 25 4
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我想构建我自己的单元测试库,我想在其中设置测试用例如下:

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及以上给出了以下错误:

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

最佳答案

改变:

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...);

关于c++ - 声明指向成员函数的可变参数模板,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32632323/

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