如何用模板参数包的内容填充数组？ [英] How to fill array with contents of a template parameter pack?
问题描述
在我发现它不符合标准之前，我已经嵌套了与VS 2015一起使用的部分专用模板代码。我希望如此，所以我扭曲了代码以克服前一个问题，并且那个现在已经遇到了困难。 / p>
使用可变参数模板和部分专业化，我想在给定固定参数集的情况下在编译时填充数组。
我想要实现的目标似乎也类似于此答案，但是我没有设法使它起作用。
p>请考虑以下程序：
#include< cstdlib>
模板< typename T，std :: size_t Size>
结构数组；
模板< typename T，std :: size_t大小，std :: size_t迭代，类型名... Args>
struct ArrayFiller {
inline
static void fill（Array< T，Size>& a，const Args& ... args）{
ArrayFiller< T，Size，Iteration ，Args ...> :: fill_recursive（a，args ...）;
}
内联
静态void fill_recursive（Array< T，Size>& a，const T& i，const Args& ... args）{
a.data [Size-Iteration-1] = i;
ArrayFiller< T，大小，迭代-1> :: fill_recursive（a，args ...）;
}
};
模板< typename T，std :: size_t Size>
结构ArrayFiller< T，大小，0> {
inline
static void fill_recursive（Array< T，Size>& a，const T& i）{
a.data [Size-1] = i;
}
};
模板< typename T，std :: size_t Size>
struct Array {
T data [Size];
模板< typename ... Args>
Array（const Args& ... args）{
ArrayFiller< T，Size，Size-1，Args ...> :: fill（* this，args ...）;
}
};
int main（）{
Array< int，2> c（42，-18）;
返回0；
}
...及其 g ++的开头- std = c ++ 14 -pedantic -Wall -Wextra 输出（从5.3.0版开始）：
main.cpp：在静态无效ArrayFiller< T，大小，迭代，Args> :: fill（Array< T，Size>& ;, const Args& ...）[带有T = int; long unsigned int大小= 2ul; long unsigned int迭代= 1ul; Args = {int，int}]'：
main.cpp：34：54：需要'Array< T，Size> :: Array（const Args& ...）[with Args = {int，int }; T = int; long unsigned int Size = 2ul]'
main.cpp：39：28：从此处需要
main.cpp：10：65：错误：没有匹配函数可用于调用'ArrayFiller< int，2ul， 1ul，int，int> :: fill_recursive（Array< int，2ul> ;, const int& ;, const int&）'
ArrayFiller< T，Size，Iteration，Args ...> :: fill_recursive（ a，args ...）;
^
main.cpp：14：17：注意：候选对象：static void ArrayFiller< T，Size，Iteration，Args> :: fill_recursive（Array< T，Size>&，const T& ;, const Args& ...）[with T = int; long unsigned int大小= 2ul; long unsigned int迭代= 1ul; Args = {int，int}]
static void fill_recursive（Array< T，Size>& a，const T& i，const Args& ... args）{
^
main .cpp：14：17：注意：候选人需要4个参数，其中3个是
基本上，编译器抱怨没有匹配功能，因为根据我的理解，参数包在我的逻辑中扩展得太快或太晚： const T& i 递归调用中的参数弄乱了扩展。
您将如何解决？
我也对替代/更好/更清洁的解决方案感兴趣。
是解决方案是否不基于您的用例可接受的模板递归？ wandbox链接
模板< typename T，std :: size_t Size>
struct Array {
T data [Size];
模板< typename ... Args>
constexpr Array（const Args& ... args）：data {args ...} {
}
};
int main（）{
Array< int，2> c（42，-18）;
assert（c.data [0] == 42）;
assert（c.data [1] == -18）;
constexpr Array< int，2> cc（42，-18）;
static_assert（cc.data [0] == 42）;
static_assert（cc.data [1] == -18）;
}
I had nested partially specialized template code working with VS 2015 until I discovered that it was not standards-compliant. I want it to be so I twisted my code to overcome the former issue and also that one and have now hit a hard wall.
Using variadic templates and partial specialization I would like to fill an array at compile-time given a fixed set of parameters.
What I want to achieve also seems similar to this answer but I did not manage to make it work.
Consider the following program:
#include <cstdlib>
template <typename T, std::size_t Size>
struct Array;
template <typename T, std::size_t Size, std::size_t Iteration, typename ...Args>
struct ArrayFiller {
inline
static void fill(Array<T, Size>& a, const Args&... args) {
ArrayFiller<T, Size, Iteration, Args...>::fill_recursive(a, args...);
}
inline
static void fill_recursive(Array<T, Size>& a, const T& i, const Args&... args) {
a.data[Size - Iteration - 1] = i;
ArrayFiller<T, Size, Iteration - 1>::fill_recursive(a, args...);
}
};
template <typename T, std::size_t Size>
struct ArrayFiller<T, Size, 0> {
inline
static void fill_recursive(Array<T, Size>& a, const T& i) {
a.data[Size - 1] = i;
}
};
template <typename T, std::size_t Size>
struct Array {
T data[Size];
template <typename ...Args>
Array(const Args&... args) {
ArrayFiller<T, Size, Size - 1, Args...>::fill(*this, args...);
}
};
int main() {
Array<int, 2> c(42, -18);
return 0;
}
...and the beginning of its g++ -std=c++14 -pedantic -Wall -Wextra output (as of version 5.3.0):
main.cpp: In instantiation of ‘static void ArrayFiller<T, Size, Iteration, Args>::fill(Array<T, Size>&, const Args& ...) [with T = int; long unsigned int Size = 2ul; long unsigned int Iteration = 1ul; Args = {int, int}]’:
main.cpp:34:54: required from ‘Array<T, Size>::Array(const Args& ...) [with Args = {int, int}; T = int; long unsigned int Size = 2ul]’
main.cpp:39:28: required from here
main.cpp:10:65: error: no matching function for call to ‘ArrayFiller<int, 2ul, 1ul, int, int>::fill_recursive(Array<int, 2ul>&, const int&, const int&)’
ArrayFiller<T, Size, Iteration, Args...>::fill_recursive(a, args...);
^
main.cpp:14:17: note: candidate: static void ArrayFiller<T, Size, Iteration, Args>::fill_recursive(Array<T, Size>&, const T&, const Args& ...) [with T = int; long unsigned int Size = 2ul; long unsigned int Iteration = 1ul; Args = {int, int}]
static void fill_recursive(Array<T, Size>& a, const T& i, const Args&... args) {
^
main.cpp:14:17: note: candidate expects 4 arguments, 3 provided
Basically the compiler complains that there is no matching function because from what I understand the parameter pack is expanded either too "soon" or too "late" in my logic: the const T& i argument in the recursive call messes up the expansion.
How would you fix it?
I am also interested in alternate / better / cleaner solutions.
Is a solution not based on template recursion acceptable in your use case? wandbox link
template <typename T, std::size_t Size>
struct Array {
T data[Size];
template <typename ...Args>
constexpr Array(const Args&... args) : data{args...} {
}
};
int main() {
Array<int, 2> c(42, -18);
assert(c.data[0] == 42);
assert(c.data[1] == -18);
constexpr Array<int, 2> cc(42, -18);
static_assert(cc.data[0] == 42);
static_assert(cc.data[1] == -18);
}
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