请帮我释放这个动态分配的数组

row = n + 1;
col = n + 1;
//used n+1 and i=-1 to avoid segmentation faults

board = malloc(row*sizeof(char *));
for(i=-1;i<row;i++)
{       
    board[i] = malloc(col*sizeof(char));
    if(board[i] == NULL)
    {
        printf("Out of memory");
            exit(EXIT_FAILURE);
    }
}

for(i=-1; i < n+1; ++i)
{
    free(board [i]);
}
free(board);

When I try to free this array in run time, my compiler goes berserk, please explain, thank you.

当我尝试在运行时释放这个数组时,我的编译器变得狂暴,请解释一下,谢谢。

2 个解决方案

#1

arrays cannot have negative index in C.

数组在C中不能有负索引。

at the line: for(i = -1; i < row; i++)

在行:for(i = -1; i ;>

I am very sure, there is an off by one error here, where free is freeing one extra block that was not malloc()ed at the end, and you must be getting a segfault error.

我非常确定,这里有一个错误,其中free释放了一个额外的块,而不是malloc()在最后,你必须得到一个段错误。

#2

malloc returns void pointer, you must cast it. Also minimum index is zero in C.

malloc返回void指针,你必须强制转换它。 C中的最小索引也为零。

    board = (char**)malloc(row*sizeof(char *));
    for(i=0;i<row;i++)
    {       
        board[i] = (char*)malloc(col*sizeof(char));
        if(board[i] == NULL)
        {
        printf("Out of memory");
            exit(EXIT_FAILURE);
        }
    }

#1