c++ - 可变参数成员函数模板是否可以有两个参数的基本情况？

Question

考虑以下代码:

class Foo {
public:
    template<typename X, typename T, typename ... Args> static void f( );
    //template<typename X, typename T> static void f( );
    template<typename X> static void f( );
};

template<typename X> void Foo::f( ) {
    cout << "Swallowed the last argument." << endl;
}

/*
template<typename X, typename T> void Foo::f( ) {
    cout << "Swallowed last two arguments." << endl;
}
 */

template<typename X, typename T, typename ... Args> void Foo::f( ) {
    cout << sizeof (T ) << endl;
    Foo::f <X, Args...>( );
}


...
    Foo::f<void*, int, double, long>( );
...

这工作正常，输出是:

4
8
8
Swallowed the last argument.

但是在我正在处理的特定问题中，我想在前一级停止递归，如注释掉的代码所示。当这些行没有被注释掉时，我得到错误(gcc 4.8.3):

g++ -Wall -std=c++1y -march=native -fmessage-length=0 -Wno-multichar   -g  -c -o obj/OneOff/OneOff.o OneOff/OneOff.cpp
OneOff/OneOff.cpp: In instantiation of ‘static void Foo::f() [with X = void*; T = double; Args = {long int}]’:
OneOff/OneOff.cpp:132:23:   required from ‘static void Foo::f() [with X = void*; T = int; Args = {double, long int}]’
OneOff/OneOff.cpp:137:36:   required from here
OneOff/OneOff.cpp:132:23: error: call of overloaded ‘f()’ is ambiguous
  Foo::f <X, Args...>( );
                       ^
OneOff/OneOff.cpp:132:23: note: candidates are:
OneOff/OneOff.cpp:130:58: note: static void Foo::f() [with X = void*; T = long int; Args = {}]
 template<typename X, typename T, typename ... Args> void Foo::f( ) {
                                                          ^
OneOff/OneOff.cpp:125:39: note: static void Foo::f() [with X = void*; T = long int]
 template<typename X, typename T> void Foo::f( ) {
                                       ^
gmake: *** [obj/OneOff/OneOff.o] Error 1

是否有可能以某种方式告诉编译器两个参数版本是特化/基本情况。例如。使用像 < .. , typename Args = {}>?

这样的东西

我发现的一种替代方法是将 f() 更改为:

template<typename X, typename T, typename ... Args> void Foo::f( ) {
    cout << sizeof (T ) << endl;
    if ( sizeof ...( Args ) > 1 ) {
        Foo::f <X, Args...>( );
    } else {
        cout << "Swallowed last two arguments." << endl;
    }
}

这给出了期望的结果，并且可能一旦生成了个体变体，“if ... elses”就会与基本情况一起被优化掉。然而，它看起来有点像 hack。

Answer 1

选项#1

让可变版本至少采用三个固定类型模板参数:

template <typename X, typename T, typename V, typename... Args>
static void f();

// ...

template <typename X, typename T, typename V, typename... Args>
void Foo::f()
{
    Foo::f<X, V, Args...>( );
}

DEMO 1

选项#2

如果包的大小等于 0，则使用 SFINAE 禁用可变版本:

#include <type_traits>

template <typename X, typename T, typename... Args>
static auto f()
    -> typename std::enable_if<(sizeof...(Args) > 0)>::type;

//...

template <typename X, typename T, typename... Args>
auto Foo::f()
    -> typename std::enable_if<(sizeof...(Args) > 0)>::type
{
    Foo::f<X, Args...>( );
}

DEMO 2