函数模板中的魔术参数 [英] Magic arguments in function templates

问题描述

在以下代码中

#include<iostream>

 template<typename T,size_t N> 
 void cal_size(T (&a)[N])
 { 
     std::cout<<"size of array is: "<<N<<std::endl;
 }

 int main()
 {
     int a[]={1,2,3,4,5,6};
     int b[]={1};

     cal_size(a);
     cal_size(b);
 }

正如预期的那样，两个数组的大小都被打印。但是如何N自动获得初始化为正确的数组大小的值（数组通过引用传递）？上述代码的工作原理是什么？

As expected the size of both the arrays gets printed. But how does N automatically gets initialized to the correct value of the array-size (arrays are being passed by reference)? How is the above code working?

推荐答案

初始化到任何东西。它不是变量。它不是一个对象。 N 是编译时常量。 N 仅在编译期间存在。 N 的值以及实际的 T 由称为模板参数扣除 em>。从传递给模板函数的参数的实际类型中推导出 T 和 N 。

N does not get "initialized" to anything. It is not a variable. It is not an object. N is a compile-time constant. N only exists during compilation. The value of N as well as the actual T is determined by the process called template argument deduction. Both T and N are deduced from the actual type of the argument you pass to your template function.

在第一次调用中，参数类型为 int [6] ，因此编译器推断 T == int 和 N == 6 ，生成一个单独的函数并调用它。让我们将它命名为 cal_size_int_6

In the first call the argument type is int[6], so the compiler deduces that T == int and N == 6, generates a separate function for that and calls it. Let's name it cal_size_int_6

void cal_size_int_6(int (&a)[6]) 
{ 
  std::cout << "size of array is: " << 6 << std::endl; 
} 

注意，没有 T 并且在此函数中不再有 N 。

Note that there's no T and no N in this function anymore. Both were replaced by their actual deduced values at compile time.

在第一次调用中，参数类型为 int [1] ，因此编译器推导出 T == int 和 N == 1 以及调用它。让我们将它命名为 cal_size_int_1

In the first call the argument type is int[1], so the compiler deduces that T == int and N == 1, generates a separate function for that as well and calls it. Let's name it cal_size_int_1

void cal_size_int_1(int (&a)[1]) 
{ 
  std::cout << "size of array is: " << 1 << std::endl; 
} 

您的主要实质上可转换为

int main() 
{ 
  int a[]={1,2,3,4,5,6}; 
  int b[]={1}; 

  cal_size_int_6(a); 
  cal_size_int_1(b); 
} 

换句话说，您的 cal_size 模板生成两个不同的函数（即原始模板的特殊化），每个都有不同的值 N （和 T ）硬编码到正文中。这就是模板在C ++中的工作原理。

In other words, your cal_size template gives birth to two different functions (so called specializations of the original template), each with different values of N (and T) hardcoded into the body. That's how templates work in C++.

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