use*_*092 45 c++ multidimensional-array
嗨,我是C++的新手,我试图从一个函数返回一个二维数组.就是这样的
int **MakeGridOfCounts(int Grid[][6])
{
int cGrid[6][6] = {{0, }, {0, }, {0, }, {0, }, {0, }, {0, }};
return cGrid;
}
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Sof*_*ner 41
此代码返回2d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;
}
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R S*_*ahu 13
使用指向指针的指针的更好替代方法是使用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));
}
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您正在(尝试做)/在您的代码段中做的是从函数返回一个局部变量,这根本不是推荐的 - 根据标准也不允许。
如果你想int[6][6]
从你的函数中创建一个,你要么必须在自由存储上为它分配内存(即使用新的 T/malloc或类似的函数),要么将已经分配的内存块传递给MakeGridOfCounts
.
该代码行不通,并且如果我们对其进行修复,也不会帮助您学习正确的C ++。最好做一些不同的事情。原始数组(尤其是多维数组)很难正确地传递到函数或从函数传递。我认为从代表数组但可以安全复制的对象开始会更好。查找有关的文档std::vector
。
在您的代码中,可以使用vector<vector<int> >
或者可以模拟具有36个元素的二维数组vector<int>
。