没有形式参数的可变函数模板 [英] variadic function template without formal parameters

问题描述

这是我想要做的：

// base case
void f() {}

template <typename T, typename... Ts>
void f() {
    // do something with T
    f<Ts...>();
}

int main() {
    f<int, float, char>();
    return 0;
}

它不编译：

prog.cpp: In instantiation of ‘void f() [with T = char; Ts = {}]’:
prog.cpp:6:5:   recursively required from ‘void f() [with T = float; Ts = {char}]’
prog.cpp:6:5:   required from ‘void f() [with T = int; Ts = {float, char}]’
prog.cpp:10:25:   required from here
prog.cpp:6:5: error: no matching function for call to ‘f()’
prog.cpp:6:5: note: candidate is:
prog.cpp:4:6: note: template<class T, class ... Ts> void f()
prog.cpp:4:6: note:   template argument deduction/substitution failed:
prog.cpp:6:5: note:   couldn't deduce template parameter ‘T’

这个线程显示了一种解决这个问题的方法，但基本案例必须是一个模板。我真的不喜欢它，因为根据我的理解，我将不得不重复代码与T一起工作。有一种方法可以避免这样吗？

This thread shows a way to fix this, but the base case has to be a template. I don't really like it, because as far as I understand I will have to duplicate code that works with T. Is there a way to avoid that?

到目前为止我想出了两个解决方案（ http://ideone.com/nPqU0l ）：

So far I came up with two solutions (http://ideone.com/nPqU0l):

template <typename...> struct types_helper {};

// base case
void f(types_helper<>) {}

template <typename T, typename... Ts>
void f(types_helper<T, Ts...>) {
    // do something with T
    f(types_helper<Ts...>());
}

int main() {
    f(types_helper<int, float, char>());
    return 0;
}

http://ideone.com/yyg6y9 ：

#include <type_traits>

struct end_of_list;

template <typename T>
void f() {
    static_assert(std::is_same<T, end_of_list>::value, "error");
}

template <typename T1, typename T2, typename... Ts>
void f() {
    // do something with T
    f<T2, Ts...>();
}

int main() {
    f<int, float, char, end_of_list>();
    return 0;
}

我不知道有没有更好的方法。

I wonder if there is a better way to do this.

推荐答案

另一种方法是将非模板函数 f 转换为可变模板函数接受零个或多个模板参数（另一个 f 需要一个或多个模板参数）。然后为了避免歧义，当参数数量不为零时，SFINAE去掉这个模板函数。好了，代码比1000个字更好：

Another way is turning the non-template function f into a variadic template function which accepts zero or more template arguments (the other f requires one or more template arguments). Then to avoid ambiguity, SFINAE away this template function when the number of arguments is not zero. Well, a code is better than 1000 words:

#include <type_traits>

template <typename... Ts>
typename std::enable_if<sizeof...(Ts) == 0>::type f() {
}

template <typename T, typename... Ts>
void f() {
    // do something with T
    f<Ts...>();
}

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