Question

嗨，我是C ++的新手我试图从函数返回一个二维数组。它是这样的

int **MakeGridOfCounts(int Grid[][6])
{
  int cGrid[6][6] = {{0, }, {0, }, {0, }, {0, }, {0, }, {0, }};

  return cGrid;
}

Answer 1

此代码返回一个二维数组。

 #include <cstdio>

    // Returns a pointer to a newly created 2d array the array2D has size [height x width]

    int** create2DArray(unsigned height, unsigned width)
    {
      int** array2D = 0;
      array2D = new int*[height];

      for (int h = 0; h < height; h++)
      {
            array2D[h] = new int[width];

            for (int w = 0; w < width; w++)
            {
                  // fill in some initial values
                  // (filling in zeros would be more logic, but this is just for the example)
                  array2D[h][w] = w + width * h;
            }
      }

      return array2D;
    }

    int main()
    {
      printf("Creating a 2D array2D\n");
      printf("\n");

      int height = 15;
      int width = 10;
      int** my2DArray = create2DArray(height, width);
      printf("Array sized [%i,%i] created.\n\n", height, width);

      // print contents of the array2D
      printf("Array contents: \n");

      for (int h = 0; h < height; h++)
      {
            for (int w = 0; w < width; w++)
            {
                  printf("%i,", my2DArray[h][w]);
            }
            printf("\n");
      }

          // important: clean up memory
          printf("\n");
          printf("Cleaning up memory...\n");
          for (  h = 0; h < height; h++)
          {
            delete [] my2DArray[h];
          }
          delete [] my2DArray;
          my2DArray = 0;
          printf("Ready.\n");

      return 0;
    }

Answer 2

你在你的片段中做什么（尝试做）/做什么是从函数中返回一个局部变量，这根本不是推荐的 - 也不是根据标准允许的。

如果您想从函数中创建int[6][6]，您必须在免费商店中为其分配内存（即使用 new T / malloc 或类似的功能），或将已经分配的内存传递给MakeGridOfCounts。

Answer 3

该代码不起作用，如果我们修复它，它不会帮助你学习正确的C ++。如果你做了不同的事情，那就更好了。原始数组（尤其是多维数组）很难正确地传递到函数和从函数传递。我认为从一个代表数组但可以安全复制的对象开始，你会好得多。查看std::vector的文档。

在您的代码中，您可以使用vector<vector<int> >，也可以使用36个元素vector<int>来模拟二维数组。

Answer 4

使用指针指针的更好选择是使用std::vector。这会处理内存分配和释放的细节。

std::vector<std::vector<int>> create2DArray(unsigned height, unsigned width)
{
   return std::vector<std::vector<int>>(height, std:vector<int>(width, 0));
}

Answer 5

#include <iostream>
using namespace std ;

typedef int (*Type)[3][3] ;

Type Demo_function( Type ); //prototype

int main (){
    cout << "\t\t!!!!!Passing and returning 2D array from function!!!!!\n"

    int array[3][3] ;
    Type recieve , ptr = &array;
    recieve = Demo_function( ptr ) ;

    for ( int i = 0 ;  i < 3 ; i ++ ){
        for ( int j = 0 ; j < 3 ; j ++ ){
            cout <<  (*recieve)[i][j] << " " ;
        }
    cout << endl ; 
    }

return 0 ;
}


Type Demo_function( Type array ){/*function definition */

    cout << "Enter values : \n" ;
    for (int i =0 ;  i < 3 ; i ++)
        for ( int j = 0 ; j < 3 ; j ++ )
            cin >> (*array)[i][j] ;

    return array ; 
}

Answer 6

你在功能上做的任何改变都会持续存在。所以你不需要返回任何东西。你可以传递2d数组并随时改变它。

  void MakeGridOfCounts(int Grid[][6])
    {
      cGrid[6][6] = {{0, }, {0, }, {0, }, {0, }, {0, }, {0, }};

    }

或

void MakeGridOfCounts(int Grid[][6],int answerArray[][6])
    {
     ....//do the changes in the array as you like they will reflect in main... 
    }

Answer 7

int** create2DArray(unsigned height, unsigned width)
{
     int** array2D = 0;
     array2D = new int*[height];

     for (int h = 0; h < height; h++)
     {
          array2D[h] = new int[width];

          for (int w = 0; w < width; w++)
          {
               // fill in some initial values
               // (filling in zeros would be more logic, but this is just for the example)
               array2D[h][w] = w + width * h;
          }
     }

     return array2D;
}

int main ()
{

    printf("Creating a 2D array2D\n");
    printf("\n");

    int height = 15;
    int width = 10;
    int** my2DArray = create2DArray(height, width);
    printf("Array sized [%i,%i] created.\n\n", height, width);

    // print contents of the array2D
    printf("Array contents: \n");

    for (int h = 0; h < height; h++)
    {
         for (int w = 0; w < width; w++)
         {
              printf("%i,", my2DArray[h][w]);
         }
         printf("\n");
    }

    return 0;
}

Answer 8

返回指向所有行的起始元素的指针数组是返回2d数组的唯一方法。

Answer 9

我建议你Matrix library作为c ++的开源工具，它的用法就像c ++中的数组一样。在这里，您可以看到documention。

Matrix funcionName(){

    Matrix<int> arr(2, 2);

    arr[0][0] = 5;
    arr[0][1] = 10;
    arr[1][0] = 0;
    arr[1][1] = 44;

    return arr;
}

Answer 10

该函数返回静态2D数组

const int N = 6;
int (*(MakeGridOfCounts)())[N] {
 static int cGrid[N][N] = {{0, }, {0, }, {0, }, {0, }, {0, }, {0, }};
 return cGrid;
}

int main() {
int (*arr)[N];
arr = MakeGridOfCounts();
}

您需要使数组静态，因为它将具有块作用域，当函数调用结束时，将创建并销毁数组。静态作用域变量持续到程序结束。

从函数返回一个二维数组

10 个答案: