C++ 使用memcpy填充动态2d数组后删除该数组时中止(内核转储) #包括 #包括 使用名称空间std; 常量字符*level1[23]= { “XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX”, “X X”, “X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX”, “X X K X D X”, “X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX X”, “X X X XXXXX X”, “X X XXXXX X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX X”, “X X X X X X X XXXXX”, “X X xxxxx X xxxxxxxxx X X X xxxxx X”, “X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X, “X XXXXXXXXXXXXXX D X XXXXK X KX”, “X X X KX XXXXXX XXXXXX XXXXXX X XXXXXX X XXXXXX DX”, “X X X X X X X X X X X X X X X D X”, “X X X X X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX X X XXXXXX X”, “X X X X X KX X X”, “X X X X X X xxxxxxxxxx X xxxxxxxxxx X xxxxxxxxxx X xxxxxx X”, “X X X X X X KX X X X X”, “X X X X X X X X X X xxxxxxxxx X”, “X X X X xxxxxxxxx X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X, “X X X X X X X X X X X X X X X”, “X X XXXXXXXXXXXXX X XXXXXXXXX X X XXXXXXXXXXXXXX X”, “X X X”, “XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX” }; //玩家结构 结构播放器信息 { int x;//x位置 int y;//y位置 int keys;//播放器拥有的键数 }; //当前游戏信息结构 结构游戏信息 { char**board;//迷宫阵列 int maxx;//迷宫数组X坐标的最大大小 int maxy;//迷宫数组X坐标的最大大小 int level;//迷宫的级别 playerinfo player;//链接到播放器数据结构 }; //初始化存储迷宫的动态数组 char**initboard(gameinfo游戏) { 字符**矩阵=新字符*[game.maxx]; 对于(int i=0;i
您通过不调用new在堆栈中分配了此数组,因此无需在最后实际删除此数组。代码运行完成后,将处理此数组!您无法在堆栈上删除,或者将从您无权访问的地址删除,因此中止。此行:C++ 使用memcpy填充动态2d数组后删除该数组时中止(内核转储) #包括 #包括 使用名称空间std; 常量字符*level1[23]= { “XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX”, “X X”, “X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX”, “X X K X D X”, “X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX X”, “X X X XXXXX X”, “X X XXXXX X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX X”, “X X X X X X X XXXXX”, “X X xxxxx X xxxxxxxxx X X X xxxxx X”, “X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X, “X XXXXXXXXXXXXXX D X XXXXK X KX”, “X X X KX XXXXXX XXXXXX XXXXXX X XXXXXX X XXXXXX DX”, “X X X X X X X X X X X X X X X D X”, “X X X X X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX X X XXXXXX X”, “X X X X X KX X X”, “X X X X X X xxxxxxxxxx X xxxxxxxxxx X xxxxxxxxxx X xxxxxx X”, “X X X X X X KX X X X X”, “X X X X X X X X X X xxxxxxxxx X”, “X X X X xxxxxxxxx X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X, “X X X X X X X X X X X X X X X”, “X X XXXXXXXXXXXXX X XXXXXXXXX X X XXXXXXXXXXXXXX X”, “X X X”, “XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX” }; //玩家结构 结构播放器信息 { int x;//x位置 int y;//y位置 int keys;//播放器拥有的键数 }; //当前游戏信息结构 结构游戏信息 { char**board;//迷宫阵列 int maxx;//迷宫数组X坐标的最大大小 int maxy;//迷宫数组X坐标的最大大小 int level;//迷宫的级别 playerinfo player;//链接到播放器数据结构 }; //初始化存储迷宫的动态数组 char**initboard(gameinfo游戏) { 字符**矩阵=新字符*[game.maxx]; 对于(int i=0;i,c++,arrays,string,pointers,C++,Arrays,String,Pointers,您通过不调用new在堆栈中分配了此数组,因此无需在最后实际删除此数组。代码运行完成后,将处理此数组!您无法在堆栈上删除,或者将从您无权访问的地址删除,因此中止。此行: #include <iostream> #include <cstring> using namespace std; const char* level1[23] = { "XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
#include <iostream>
#include <cstring>
using namespace std;
const char* level1[23] =
{
"XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX",
"X X",
"X XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX X",
"X X K X D X X",
"X XXXXXXXXXXXXXXXXXXXXXX XXXXXXXXX XXXXXXXXXXXXXXXXXXXXX X X",
"X X X X X XXXXX X X",
"X X XXXXX X XXXXXXXXXXXXXX XXXXXXXX XXXXXXXXXXX X X",
"X X X X D X X X X X XXXXX X",
"X X XXXX X X XXXXXXXXXX X X X XXXX X X",
"X X X X E X XXXXXXXXX XXXXXXXX X XXXXX X X",
"X XXXXXXXXXXXXXXXX D X XXK X X X X KX X",
"X X X X KX XXXXXX XXXXXX XXXXXXXX X X X XXXXXXDX",
"X X X X X X X XXX X X X X D X X",
"X X X X X X XXXXXXXXXXXXXXX XXXXXXXXXXXXXX X XXXXX X X",
"X X X X X X X X X KX X X X",
"X X X X X X X X XXXXXXXXXXX X XXXXXXXXXXX X XXXXX X",
"X X X X X X X X X X KX X X X X X",
"X X X X X X X X XX X X XXXXXXXXXXXXX X X",
"X X X X X X X X X XXXXXXXX X X X X X X",
"X X X X X X X X X X X X X X",
"X X XXXXXXXXXXXXX X XXXXXXXX X X X XXXXXXXXXXXXXXXXXX X",
"XS X X",
"XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX"
};
// Player structure
struct playerinfo
{
int x; // X position
int y; // Y position
int keys; // Number of keys the player has
};
// Current game info structure
struct gameinfo
{
char** board; // The maze array
int maxx; // Max size of the maze array X coordinate
int maxy; // Max size of the maze array X coordinate
int level; // The level of the maze
playerinfo player; // Link to player data struct
};
// Initializes the dynamic array that stores the maze
char** initboard(gameinfo game)
{
char** matrix = new char*[game.maxx];
for (int i=0; i<game.maxx; i++)
matrix[i] = new char[game.maxy];
return matrix;
}
void delboard(gameinfo game)
{
for (int i=0;i<game.maxx;i++)
delete [] game.board[i];
delete [] game.board;
}
int main()
{
gameinfo game;
game.maxx=65;
game.maxy=23;
game.board = initboard(game);
memcpy(game.board,level1,sizeof(level1));
delboard(game);
}
这是错误的,因为:
game.board
的内容,该游戏板被level1
的内容分配用来保存char*
数组initboard
将level1
的内容复制到正确的内存位置并删除上面的行来解决此问题
memcpy(game.board,level1,sizeof(level1));
然后,您需要在initboard
中分配game.maxy+1
字符
game.maxx=23; // Not 65;
game.maxy=65 // Not 23;
否则,在
matrix[i]
中将char**board;
更改为vector board;
中,将没有足够的空间容纳终止的空字符。清除initboard
和delboard
,并在main()
中:
matrix[i] = new char[game.maxy+1];
game.board=vector(开始(级别1),结束(级别1));
您可能需要包含向量
,字符串
,迭代器
。您还可以删除maxx
和maxy
并直接从板
读取它们
…看看这一切有多简单?你可能想问问自己,当你调用它时,你实际上在
memcpy
'做什么?你正在对刚刚分配的整组行缓冲区指针进行轰炸,然后试图删除[]
大量指针从未实际分配过。更糟糕的是,您这样做了两次,每次都会泄漏整个动态行缓冲区集。这被标记为C。为什么,确切地说?这根本不可能“完美工作”您正在用静态表中的const char*
序列替换动态行缓冲区指针序列,然后调用delete[]在这些指针上,编译器后端并不重要,你所做的是UB。这确实让我感到震惊,因为我尝试用C++编写C代码。使用C++容器中的一些C++习语,我保证你的生活会变得简单很多。注意:OP的索引使用不正确,在调用这个之前需要加以解决。level1
数组是指向字符的23个常量指针,每个指针指向一行65+1个字符宽(包括终止符)。传递给此的game
中的值是这样的:它将分配一个包含65个字符指针的数组,然后为每个指针分配一个23个元素宽的字符缓冲区,通过将65+1长度的字符串复制到23个字符的空间中来调用UB,并为过程中的每一行调用UB。@WhozCraig感谢您注意到问题并让我知道w、 我已经更新了答案。level1
不是“在堆栈上分配的”!我真的很喜欢它的工作方式,我需要学习更多关于vector的知识以及如何在将来使用它们。我能够在没有它们的情况下使它工作,但对于任何未来的项目来说,它看起来都是一种更好的方法。
matrix[i] = new char[game.maxy+1];
game.board = vector<string>( begin(level1), end(level1) );