C ++模板参数扣除的std ::阵列与非为size_t整数 [英] C++ template parameter deduction for std::array with non size_t integer

问题描述

我试图适应的解决方案presented在<一个href=\"http://stackoverflow.com/questions/21723799/avoiding-struct-in-variadic-template-function\">Avoiding在结构可变参数模板函数，以我的需要。但是，我无法理解G ++的行为。考虑下面的函数：

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 });

不匹配GCC与

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>’

然而，它匹配，如果我修改无符号的大小到无符号长大小或为size_t 。我不知道，了解这里发生了什么。是不是尺寸参数调用的std ::阵列＆LT; T，尺寸和GT; 转换为为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

的std :: 是模板为数组：

std::array is templated as:

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

而大小 N 要求是类型为size_t 。但是，在你的函数，你传递一个无符号（INT），它不能作为PTED 为size_t 间$ P $。据 SFINAE 如果不能扣除的模板，它不存在，因此你的模板功能不存在的。

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.

这是不是与调用线的问题，而是你对你的函数模板的声明。要解决此问题，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;
 }

在此情况下，即使在使用

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，并铸造。

It still works because it can be deducted and casted.

由DYP附加说明：

<强> [temp.deduct.type] / 17用于非类型模板参数需要的推导的东西（模板参数）的类型是相同的类型为模板参数的它推导的。

[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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