在 CPP 的构造函数中初始化二维数组 class

Initialize 2D array in constructor of CPP class

我想知道在 cpp class 中初始化二维数组的最佳方法是什么。在调用构造函数之前我不知道它的大小,即

头文件包含:

private:
    int size;
    bool* visited;
    int edges;
    int** matrix;

默认构造函数(现在):

Digraph::Digraph(int n) {
  int rows = (n * (n-1)/2);
  int columns = 2;

  matrix = new int[rows][2];

  visited[size] = { 0 };

  size = n;
  edges = 0;
}

我要的是N行2列的二维数组

当我尝试编译时,目前 returns error: cannot convert 'int (*)[2]' to 'int**' in assignment

注意:我不会使用矢量,所以请不要推荐它们。

matrix = new int[rows][2]; 是无效语法。分配一个二维稀疏数组需要多次new[]调用,例如:

private:
    int size;
    bool* visited;
    int edges;
    int** matrix;
    int rows;
    int columns;

...

Digraph::Digraph(int n) {
  size = n;
  edges = 0;

  rows = (n * (n-1)/2);
  columns = 2;

  matrix = new int*[rows];
  for(int x = 0; x < rows; ++x) {
    matrix[x] = new int[columns];
    for(int y = 0; y < columns; ++y)
      matrix[x][y] = 0;
  }

  visited = new bool[size];
  for(int x = 0; x < size; ++x)
    visited[x] = false;
}

Digraph::~Digraph() {
  for(int x = 0; x < rows; ++x) {
    delete[] matrix[x];
  }
  delete[] matrix;

  delete[] visited;
}

或者,考虑将 matrix 分配为一维数组,然后在访问其值时使用二维索引,例如:

private:
    int size;
    bool* visited;
    int edges;
    int* matrix; // <-- 1 *, not 2 **
    int rows;
    int columns;

    int& matrix_value(int row, int col) { return matrix[(row * rows) + col]; }

...

Digraph::Digraph(int n) {
  size = n;
  edges = 0;

  rows = (n * (n-1)/2);
  columns = 2;

  n = rows * columns;
  matrix = new int[n];
  for(int x = 0; x < n; ++x)
    matrix[n] = 0;

  visited = new bool[size];
  for(int x = 0; x < size; ++x)
    visited[x] = false;
}

Digraph::~Digraph() {
  delete[] matrix;
  delete[] visited;
}

无论哪种方式,您还需要实现(或禁用)复制构造函数和复制赋值运算符,最好是移动构造函数和移动赋值运算符,根据 Rule of 3/5/0,例如:

Digraph::Digraph(const Digraph &src) {
  size = src.size;
  edges = src.edges;

  rows = src.rows;
  columns = src.columns;

  matrix = new int*[rows];
  for(int x = 0; x < rows; ++x) {
    matrix[x] = new int[columns];
    for (int y = 0; y < columns; ++y)
      matrix[x][y] = src.matrix[x][y];
  }

  /* or:
  n = rows * columns;
  matrix = new int[n];
  for(int x = 0; x < n; ++x)
    matrix[n] = src.matrix[n];
  */

  visited = new bool[size];
  for(int x = 0; x < size; ++x)
    visited[x] = src.visited[x];
}

Digraph::Digraph(Digraph &&src) {
  size = 0;
  edges = 0;
  rows = 0;
  columns = 0;
  matrix = nullptr;
  visited = nullptr;

  src.swap(*this);
}

void Digraph::swap(Digraph &other) {
  std::swap(size, other.size);
  std::swap(edges, other.edges);
  std::swap(rows, other.rows);
  std::swap(columns, other.columns);
  std::swap(matrix, src.matrix);
  std::swap(visited, src.visited);
}

Digraph& Digraph::operator=(Digraph rhs) {
    Digraph temp(std::move(rhs));
    temp.swap(*this);
    return this;
}

也就是说,更好的设计是使用 std::vector 而不是 new[],并让它为您处理所有内存管理和 copying/moving,例如:

#include <vector>

private:
    int size;
    std::vector<bool> visited;
    int edges;
    std::vector<std::vector<int>> matrix;
    // or: std::vector<int> matrix;
    int rows;
    int columns;

...

Digraph::Digraph(int n) {
  size = n;
  edges = 0;

  rows = (n * (n-1)/2);
  columns = 2;

  matrix.resize(rows);
  for(int x = 0; x < rows; ++x)
      matrix[x].resize(columns);

  /* or:
  matrix.resize(rows * columns);
  */

  visited.resize(size);
}

// implicitly-generated copy/move constructors, copy/move assignment operators,
// and destructor will suffice, so no need to implement them manually...

如果您不能使用 std::vector,请考虑使用适当的语义实现您自己的 vector class,然后改用它。你真的应该努力尽可能多地遵循 0 规则,通过使用为你实现 3/5 规则的 classes。