传递大括号初始化列表参数时调用可变函数模板的问题

Question

考虑这个函数模板：

template <class... T>
void foo (std::tuple<T, char, double> ... x);

此调用有效：

using K = std::tuple<int, char, double>;
foo ( K{1,'2',3.0}, K{4,'5',6.0}, K{7,'8',9.0} );

这个没有：

foo ( {1,'2',3.0}, {4,'5',6.0}, {7,'8',9.0} );

（gcc 和 clang 都抱怨 foo 的参数太多）

为什么第二次调用有问题？我可以重写 foo 的声明以便第二个调用也被接受吗？

模板参数 T 仅用于实现可变性。实际类型是已知且固定的，只是参数的数量不同。在现实生活中，类型与 int, char, double 不同，这只是一个例子。

我不能为此使用 C++17。 C++11 兼容的解决方案是首选。

Answer 1

{} 不是表达式因此没有类型，参数推导与类型有关，当用于执行参数推导的参数是 初始化器列表时要特别小心=38=] 模板函数参数必须有特定形式，否则参数是非推导上下文。一个更简单的例子是这样的：

template <class T> struct A { T r; }; template <class T> void foo (A<T> x); using K = A<int>; foo({1}); // fail foo(K{1}); // compile

[temp.deduc.call]/1
涵盖了这一点

If removing references and cv-qualifiers from P gives std::initializer_list<P'> or P'[N] for some P' and N and the argument is a non-empty initializer list ([dcl.init.list]), then deduction is performed instead for each element of the initializer list, taking P' as a function template parameter type and the initializer element as its argument, and in the P'[N] case, if N is a non-type template parameter, N is deduced from the length of the initializer list. Otherwise, an initializer list argument causes the parameter to be considered a non-deduced context

和[temp.deduct.type]/5

The non-deduced contexts are:

(5.6) A function parameter for which the associated argument is an initializer list ([dcl.init.list]) but the parameter does not have a type for which deduction from an initializer list is specified ([temp.deduct.call]).

当你：

显式提供模板参数，有效...没有什么可推断的

将参数指定为 K{1}，有效...参数不再是 初始化列表 ，是一个类型为
的表达式

Answer 2

生成一组重载的构造函数：

#include <tuple>
#include <cstddef>

template <typename T, std::size_t M>
using indexed = T;

template <typename T, std::size_t M, std::size_t... Is>
struct initializer : initializer<T, M, sizeof...(Is) + 1, Is...>
{    
    using initializer<T, M, sizeof...(Is) + 1, Is...>::initializer;

    initializer(indexed<T, Is>... ts)
    {
        // ts is a pack of std::tuple<int, char, double>
    }
};

template <typename T, std::size_t M, std::size_t... Is>
struct initializer<T, M, M, Is...> {};

using foo = initializer<std::tuple<int, char, double>, 20>;
//                                   tuples limit+1 ~~~^

int main()
{
    foo({1,'2',3.0});
    foo({1,'2',3.0}, {4,'5',6.0});
    foo({1,'2',3.0}, {4,'5',6.0}, {7,'8',9.0});
}

DEMO

生成一组重载的函数调用运算符：

#include <tuple>
#include <cstddef>

template <typename T, std::size_t M>
using indexed = T;

template <typename T, std::size_t M, std::size_t... Is>
struct initializer : initializer<T, M, sizeof...(Is) + 1, Is...>
{    
    using initializer<T, M, sizeof...(Is) + 1, Is...>::operator();

    int operator()(indexed<T, Is>... ts) const
    {            
        // ts is a pack of std::tuple<int, char, double>
        return 1;
    }
};

template <typename T, std::size_t M, std::size_t... Is>
struct initializer<T, M, M, Is...>
{
    int operator()() const { return 0; }
};

static constexpr initializer<std::tuple<int, char, double>, 20> foo = {};
//                                        tuples limit+1 ~~~^

int main()
{    
    foo({1,'2',3.0});
    foo({1,'2',3.0}, {4,'5',6.0});
    foo({1,'2',3.0}, {4,'5',6.0}, {7,'8',9.0});
}

DEMO 2

创建（或使用预处理器宏生成）一组将参数转发给单个实现的重载：

#include <array>
#include <tuple>

using K = std::tuple<int, char, double>;

void foo(const std::array<K*, 5>& a)
{
    // a is an array of at most 5 non-null std::tuple<int, char, double>*
}

void foo(K p0) { foo({&p0}); }
void foo(K p0, K p1) { foo({&p0, &p1}); }
void foo(K p0, K p1, K p2) { foo({&p0, &p1, &p2}); }
void foo(K p0, K p1, K p2, K p3) { foo({&p0, &p1, &p2, &p3}); }
void foo(K p0, K p1, K p2, K p3, K p4) { foo({&p0, &p1, &p2, &p3, &p4}); }

int main()
{
    foo({1,'2',3.0});
    foo({1,'2',3.0}, {4,'5',6.0});
    foo({1,'2',3.0}, {4,'5',6.0}, {7,'8',9.0});
}

DEMO 3

作为数组传递并推断其大小（需要额外的一对括号）：

#include <tuple>
#include <cstddef>

template <std::size_t N>
void foo(const std::tuple<int, char, double> (&a)[N])
{
    // a is an array of exactly N std::tuple<int, char, double>
}

int main()
{
    foo({{1,'2',3.0}, {4,'5',6.0}});
 //     ^~~~~~ extra parens ~~~~~^
}

DEMO 4

使用 std::initializer_list 作为构造函数参数（跳过额外的括号）：

#include <tuple>
#include <initializer_list>

struct foo
{
    foo(std::initializer_list<std::tuple<int, char, double>> li)
    {
        // li is an initializer list of std::tuple<int, char, double>
    }
};

int main()
{
    foo{ {1,'2',3.0}, {4,'5',6.0} };
}

DEMO 5

Answer 3

I cannot use C++17 for this. A C++11-compatible solution is much preferred.

使用 C++11 稍微复杂一点（没有 std::index_sequence，没有 std::make_index_sequence）但是，如果你想保持元组的可变参数使用......那就是......如果你非常想要

foo (std::tuple<int, char, double> ... ts)

并且如果您接受调用模板结构的静态方法，您可以定义一个递归继承自身的模板结构，并递归地定义一个

func ();
func (K t0);
func (K t0, K t1);
func (K t0, K t1, K t2);

其中 K 是你的

using K = std::tuple<int, char, double>;

下面是一个完整的编译C++11的例子

#include <tuple>
#include <iostream>

using K = std::tuple<int, char, double>;

template <typename T, std::size_t>
struct getTypeStruct
 { using type = T; };

template <typename T, std::size_t N>
using getType = typename getTypeStruct<T, N>::type;

template <int ...>
struct iList;

template <std::size_t = 50u, std::size_t = 0u, typename = iList<>>
struct foo;

template <std::size_t Top, std::size_t N, int ... Is>
struct foo<Top, N, iList<Is...>> : public foo<Top, N+1u, iList<0, Is...>>
 {
   using foo<Top, N+1u, iList<0, Is...>>::func;

   static void func (getType<K, Is> ... ts)
    { std::cout << sizeof...(ts) << std::endl; }
 };

template <std::size_t Top, int ... Is>
struct foo<Top, Top, iList<Is...>>
 {
   // fake func, for recursion ground case
   static void func ()
    { }
 };


int main()
 {
   foo<>::func({1,'2',3.0}); // print 1
   foo<>::func({1,'2',3.0}, {4,'5',6.0}); // print 2
   foo<>::func({1,'2',3.0}, {4,'5',6.0}, {7,'8',9.0});  // print 3
 }

如果你可以使用 C++14，你可以使用 std::make_index_sequence 和 std::index_sequence 并且代码会变得更好一些，恕我直言

#include <tuple>
#include <iostream>
#include <type_traits>

using K = std::tuple<int, char, double>;

template <std::size_t ... Is>
constexpr auto getIndexSequence (std::index_sequence<Is...> is)
   -> decltype(is);

template <std::size_t N>
using IndSeqFrom = decltype(getIndexSequence(std::make_index_sequence<N>{}));

template <typename T, std::size_t>
struct getTypeStruct
 { using type = T; };

template <typename T, std::size_t N>
using getType = typename getTypeStruct<T, N>::type;

template <std::size_t N = 50, typename = IndSeqFrom<N>>
struct foo;

template <std::size_t N, std::size_t ... Is>
struct foo<N, std::index_sequence<Is...>> : public foo<N-1u>
 {
   using foo<N-1u>::func;

   static void func (getType<K, Is> ... ts)
    { std::cout << sizeof...(ts) << std::endl; }
 };

template <>
struct foo<0, std::index_sequence<>>
 {
   static void func ()
    { std::cout << "0" << std::endl; }
 };

int main()
 {
   foo<>::func({1,'2',3.0});  // print 1
   foo<>::func({1,'2',3.0}, {4,'5',6.0});  // print 2
   foo<>::func({1,'2',3.0}, {4,'5',6.0}, {7,'8',9.0});  // print 3
 }

遗憾的是您不能使用 C++17，因为您可以使用可变参数 unsing 并完全避免递归继承

#include <tuple>
#include <iostream>
#include <type_traits>

using K = std::tuple<int, char, double>;

template <std::size_t ... Is>
constexpr auto getIndexSequence (std::index_sequence<Is...> is)
   -> decltype(is);

template <std::size_t N>
using IndSeqFrom = decltype(getIndexSequence(std::make_index_sequence<N>{}));

template <typename T, std::size_t>
struct getTypeStruct
 { using type = T; };

template <typename T, std::size_t N>
using getType = typename getTypeStruct<T, N>::type;

template <std::size_t N, typename = IndSeqFrom<N>>
struct bar;

template <std::size_t N, std::size_t ... Is>
struct bar<N, std::index_sequence<Is...>>
 {
   static void func (getType<K, Is> ... ts)
    { std::cout << sizeof...(ts) << std::endl; }
 };

template <std::size_t N = 50, typename = IndSeqFrom<N>>
struct foo;

template <std::size_t N, std::size_t ... Is>
struct foo<N, std::index_sequence<Is...>> : public bar<Is>...
 { using bar<Is>::func...; };

int main()
 {
   foo<>::func({1,'2',3.0});  // print 1
   foo<>::func({1,'2',3.0}, {4,'5',6.0});  // print 2
   foo<>::func({1,'2',3.0}, {4,'5',6.0}, {7,'8',9.0});  // print 3
 }

传递大括号初始化列表参数时调用可变函数模板的问题

Problem with calling a variadic function template when passing brace initialiser list arguments

c++

c++11

variadic-templates

list-initialization