Question

问题几乎没有意义。因此,我试图这样做。

总的来说,C++允许:

template<class T, class U, T t, U u>
void func() {}

func<char, int,  A , 10>();

但是,似乎自然的理论延伸并不奏效。

template<class...T, T... t>
void func() {}

func<char, int,  A , 10>();

Both clang and g++4.7 reject the above code. The error is shown where the instantiation is done. It appears to me that the two variadic lists should be parsed unambiguously because the first one has types and the other one has integral values only.

If the above is not meant to work, I think the following won t work either.

template <class Ret, class... Args, Ret (*func)(Args...)>
class Foo {};

我认为,“Foo”模板是一个非常有用的东西。

Answer 1

(Extra: 直接回答你的第一个问题,你还可以将<条码>template<}...T. t> void func(){}变成一个模板,在边段。这在++4.6中并不奏效,但在3.0部族中则如此。

在模板中添加一个模板:

template<class ... T>
struct func_types {
   template <T ... t>
   static void func_values() {
       // This next line is just a demonstration, and 
       // would need to be changed for other types:
       printf("%c %d
", t...);
   }
};

int main() {
    func_types<char, int> :: func_values< A , 10>();
}

是否可接受一个模版。另一种选择是使用带<代码>func< tuple<char,int> ,_tuple(A,10) >的标记。我认为这一点是可行的,但你可能不得不推出自己的补习班(这似乎使读本为constexpr)。

Finally, you might be able to implement your Foo template as follows:

template<typename Ret, typename ...Args>
struct Foo {
        template< Ret (*func)(Args...)>
        struct Bar {
                template<typename T...>
                Bar(T&&... args) {
                        cout << "executing the function gives: "
                           << func(std::forward(args)...) << endl;
                }
        };
};

int main () {
    Foo<size_t, const char*> ::  Bar<strlen> test1("hi");
    Foo<int, const char*, const char*> ::  Bar<strcmp> test2("compare","these");
}

This latter code is on ideone. To demonstrate, I implemented a constructor to forward the args to the function that s code into the template.

Answer 2

因为没有人会认为应该有这一特点。理论模板的设计意在简单易行。其他可能先进和有用的特征也包括在内。

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