使用非size_t整数的std :: array的C ++模板参数推导 [英] C++ template parameter deduction for std::array with non size_t integer

问题描述

我正在尝试修改避免可变模板函数中的结构到我的需要。但是，我不能理解G ++的行为。考虑以下函数：

I'm trying to adapt the solution presented in Avoiding struct in variadic template function to my need. However, I can't understand the the behavior of G++. Consider the following function:

 template <typename T, unsigned Size>
 int nextline(const typename std::array<T, Size> ar) {
    return 0;
 }

然后调用

 nextline(std::array<int, 2> { 1,0 });

与GCC抱怨不符

eslong.cpp: In function ‘int main()’:
eslong.cpp:10:38: error: no matching function for call to ‘nextline(std::array<int, 2ul>)’
   nextline(std::array<int, 2> { 1,0 });
                                      ^
eslong.cpp:10:38: note: candidate is:
eslong.cpp:4:5: note: template<class T, unsigned int Size> int nextline(std::array<T, Size>)
 int nextline(const typename std::array<T, Size> ar) {
     ^
eslong.cpp:4:5: note:   template argument deduction/substitution failed:
eslong.cpp:10:38: note:   mismatched types ‘unsigned int’ and ‘#‘integer_cst’ not supported by dump_type#<type error>’
   nextline(std::array<int, 2> { 1,0 });
                                      ^
eslong.cpp:10:38: note:   ‘std::array<int, 2ul>’ is not derived from ‘std::array<T, Size>’

但是，如果我将 unsigned Size 更改为 unsigned long Size 或 size_t 。我不知道这里发生了什么。不是调用 std :: array< T，Size> 的参数转换为 size_t ？

However it matches if I changes unsigned Size to unsigned long Size or size_t. I'm not sure to understand what's happening here. Isn't the Size parameter in the call to std::array<T, Size> converted to a size_t ?

推荐答案

std :: array 模板为：

template<class T, std::size_t N > struct array;

，而大小 N 类型 size_t 。但在你的函数中，你传递一个unsigned（int），不能被解释为 size_t 。根据 SFINAE 如果模板无法扣除，则它不存在，因此您的模板函数不存在。

while the size N is required to be the type size_t. But in your function, you are passing an unsigned (int) which cannot be interpreted as size_t. According to SFINAE If a template cannot be deducted, it does not exist, thus your templated function does not exist.

这不是调用行的问题，而是你的函数模板的声明。要更正此问题，请使用正确的类型：

It is NOT the problem with the call line, but your declaration of your function template. To correct this, use the correct type:

template <typename T, size_t Size>
int nextline(const typename std::array<T, Size> ar) {
  return 0;
 }

在这种情况下，即使您使用：

In this case, even you use:

nextline(std::array<int, 2ul> { 1,0 });

它仍然有效，因为它可以被扣除和强制转换。

It still works because it can be deducted and casted.

dyp的其他说明：

[temp.deduct .type] / 17 ，用于需要推导出的事物（模板参数）的类型与其推导的模板参数类型相同的非类型模板参数。

[temp.deduct.type]/17 for non-type template parameters that requires the type of the deduced thing (template-argument) to be of the same type as the template-parameter it is deduced for.

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