如何通过引用将数组传递给功能模板 [英] How to pass array to function template with reference

问题描述

我正在学习c ++模板概念。我不理解以下内容。

I am learning c++ template concepts. I do not understand the following.

#include <iostream>
#include <typeinfo>

using namespace std;

template <typename T>
T fun(T& x)
{
 cout <<" X is "<<x;
 cout <<"Type id is "<<typeid(x).name()<<endl;
}


int main ( int argc, char ** argv)
{
   int a[100];
   fun (a);
}

我正在尝试什么？

1）T fun（T& x）

1) T fun (T & x)

这里x是一个引用，因此不会将'a'衰减为指针类型
但是在编译时，出现以下错误。

Here x is a reference, and hence will not decayed 'a' into pointer type, but while compiling , i am getting the following error.

 error: no matching function for call to ‘fun(int [100])’

当我尝试非引用时，它可以正常工作。据我了解，数组被分解为指针类型。

When I try non-reference, it works fine. As I understand it the array is decayed into pointer type.

推荐答案

C样式数组是非常基本的构造，不能以内置或用户定义的方式进行分配，复制或引用类型是。要实现通过引用传递数组的等效功能，您需要以下语法：

C-style arrays are very basic constructs which are not assignable, copyable or referenceable in the way built-ins or user defined types are. To achieve the equivalent of passing an array by reference, you need the following syntax:

// non-const version
template <typename T, size_t N>
void fun( T (&x)[N] ) { ... }

// const version
template <typename T, size_t N>
void fun( const T (&x)[N] ) { ... }

请注意，这里的数组大小也是模板参数，允许函数使用所有数组大小，因为 T [M] 和 M ， N ，> T [N] 的类型不同。另请注意，该函数返回void。如前所述，由于数组不可复制，因此无法按值返回数组。

Note that here the size of the array is also a template parameter to allow the function to work will all array sizes, since T[M] and T[N] are not the same type for different M, N. Also note that the function returns void. There is no way of returning an array by value, since the array is not copyable, as already mentioned.

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