如何模板函数“知道”作为模板参数给定的数组的大小？ [英] How can a template function 'know' the size of the array given as template argument?

问题描述

在下面的C ++代码中，模板化的Check函数给出了一个不是我想要的输出：它是1而不是3.我怀疑K映射到 int * ，而不是 int [3] （是类型？）。我想它给我相同的输出比第二个（非模板）函数，我明确给出的数组的大小...

使用短宏，有没有办法写一个接受单个参数但仍知道数组大小的Check函数。

 ＃ include< iostream> 
 using namespace std; 
 
 int data [] = {1,2,3}; 
 
 template< class K> 
 void Check（K data）{
 cout<< 推导的大小：< sizeof（data）/ sizeof（int）< endl; 
} 
 
 void Check（int *，int sizeofData）{
 cout< 正确大小：< sizeofData / sizeof（int）< endl; 
} 
 
 int main（）{
 Check（data）; 
检查（data，sizeof（data））; 
} 
  

感谢。

PS：在实际代码中，数组是一个结构体数组，在进行单元测试时必须重复。

解决方案

p> template< class T，size_t S>
void Check（T（&）[S]）{
cout<< 推导的大小：< S<< endl;
}


In the C++ code below, the templated Check function gives an output that is not what I would like: it's 1 instead of 3. I suspect that K is mapped to int*, not to int[3] (is that a type?). I would like it to give me the same output than the second (non templated) function, to which I explicitly give the size of the array...

Short of using macros, is there a way to write a Check function that accepts a single argument but still knows the size of the array?

#include <iostream>
using namespace std;

int data[] = {1,2,3};

template <class K>
void Check(K data) {
  cout << "Deduced size: " << sizeof(data)/sizeof(int) << endl;
}

void Check(int*, int sizeofData) {
  cout << "Correct size: " << sizeofData/sizeof(int) << endl;
}

int main() {
  Check(data);
  Check(data, sizeof(data));
}

Thanks.

PS: In the real code, the array is an array of structs that must be iterated upon for unit tests.

解决方案

template<class T, size_t S> 
void Check(T (&)[S]) {  
   cout << "Deduced size: " << S << endl;
}

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