将类级别的 2D 数组变量设置为传入值 - C++ [英] Setting class level 2D Array variable to passed in value - C++

问题描述

我正在尝试制作一个解谜程序，虽然算法是合理的(至少在我的脑海中)，但我遇到了二维数组的障碍.我来自 C# 和 Java，所以语法让我很伤心.

I'm attempting to make a maze-solving program, and while the algorithm is sound (at least in my head), I've been running into a roadblock with 2D arrays. I'm coming from C# and Java, so the syntax is giving me grief.

这是一个 SSCCE:

Here's an SSCCE:

//Main.cpp

#include "MazeSolver.h"

int main()
{
     MazeSolver mazeSolver;

     char maze[51][51] = { }; // Not bothering to show code for populating the array

     mazeSolver.setMaze(maze);
}

<小时>

//MazeSolver.cpp

#include "MazeSolver.h"

MazeSolver::MazeSolver() { }

void MazeSolver::setMaze(char maze[51][51])
{
     this->maze = maze;
}

<小时>

//MazeSolver.h

#ifndef _MAZESOLVER_H
#define _MAZESOLVER_H

class MazeSolver
{
private:
     char **maze; // This is likely the problem, I'm at wits end in regards to
                  // what notation is needed here.  I do not want to hard-copy
                  // the entire array, I just need to access it (via pointer)
                  // I've also tried just using char maze[51][51] here, char *maze, etc...
public:
      MazeSolver();
      void setMaze(char maze[51][51]);
}

推荐答案

这是 C++ 的怪癖之一.

This is one of the weird quirks of C++.

C++ 2D 数组不是锯齿状数组.当你声明 char maze[51][51] 时，它实际上分配了 1 个 51*51 成员长度的连续数组.sizeof(迷宫) == 51*51.当你取消引用一个值时，maze[a][b]，它实际上做的是*(maze+51*a+b).所有这些都在幕后.

C++ 2D arrays are NOT jagged arrays. When you declare char maze[51][51], it actually allocates 1 contiguous array 51*51 members long. sizeof(maze) == 51*51. When you dereference a value, maze[a][b], what it actually does is *(maze+51*a+b). All this is under the hood.

锯齿状数组是一个数组数组，一个字符**.在这种情况下，您有一个包含 51 个指针大小 == (51*sizeof(void*)) 的数组.在每个位置，指针指向一个完全不同的内存位置，分配给51个成员长.

A Jagged Array is an array of arrays, a char**. In this case, you have an array of 51 pointers size == (51*sizeof(void*)). In each position, the pointer points to a completely different memory location, allocated to 51 members long.

这很烦人，因为您不能只转换两者，即使是强制转换也是如此.您必须处理奇怪的语法，例如 char (*maze)[51] 来获取指向二维数组的指针.

This is ANNOYING because you can't just convert the two, even by casting. You have to deal with weird syntax, such as char (*maze)[51] to get a pointer to the 2D array.

更烦人的是以下情况:

int foo(int maze[51][51])
{
   return sizeof(maze);
}

int maze[51][51];

int main(int argc, char** argv)
{
   std::cout << sizeof(maze) << std::endl;//return 51*51*sizeof(int);
   std::cout << foo(maze) << std::endl;//return 8, sizeof(void*);
}

所以它隐式地按引用传递，而不是按值传递，这与 C++ 的所有其余部分相反.

So it implicitly passes by reference, not by value, which is opposite all the rest of C++.

tl;博士;

指向二维数组的指针的正确语法是char (*maze)[51];.您的语法用于锯齿状数组(数组数组)，这在 C++ 中是不同的.

The correct syntax for a pointer to a 2D array is char (*maze)[51];. Your syntax is for a jagged array (arrays of arrays) which is NOT the same thing in C++.

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