Question

问题是数组的声明。

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#include <cstdlib>
#include <fstream>
#include <iostream>
#include <vector>
#include <list> 

template <typename Type, size_t const SIZE>
class dummy_array {
    Type data[SIZE] = {};

  public:
    dummy_array(){}
    ~dummy_array(){}

    Type& operator[](size_t const index)
    {
      if (index < SIZE)
        return data[index];
      throw std::out_of_range("index out of range");
    }

    Type const& operator[](size_t const index) const
    {
      if (index < SIZE)
        return data[index];
      throw std::out_of_range("index out of range");
    }

};
int main()
{
    {
      dummy_array<int, 6> arr();
      arr[0] = 1;
      arr[1] = 2;

      for (int i = 0; i < 6; i++)
        std::cout << arr[i] << " " ;
    std::cout << std::endl;

    }
    return 0;
}

有人可以解释为什么用“ dummy_array arr（）;”声明吗？导致失败如下。构建日志：

main.cpp: In function 'int main()':

main.cpp:34:12: error: pointer to a function used in arithmetic [-Wpointer-arith]

   arr[0] = 1;

        ^

main.cpp：34：16：错误：分配了只读位置'* arr'

   arr[0] = 1;

            ^

main.cpp：35：12：错误：指向算术[-Wpointer-arith]中使用的函数的指针

   arr[1] = 2;

        ^

main.cpp：35：16：错误：分配了只读位置'*（arr + 1）'

   arr[1] = 2;

            ^

main.cpp：38：27：错误：指向算术[-Wpointer-arith]中使用的函数的指针

     std::cout << arr[i] << " " ;

                       ^

Answer 1

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这可以看作是函数声明或变量声明。

编译器始终喜欢使用函数声明而不是变量声明，因此实际上您声明了一个名为arr的函数，不带任何参数并返回dummy_array。改用它：

dummy_array<int, 6> arr();

因dummy_array <int，6> arr（）;

1 个答案: