Question

我有一个包含unsigned char *的结构，用于存储任意数据。在某些时候，我想使用这些数据，就好像它是一个二维数组。我就是这样做的：

#define DATA_SIZE 10
unsigned char *data = malloc(DATA_SIZE * DATA_SIZE * sizeof(unsigned char));

// this is not allowed
unsigned char (* matrix)[DATA_SIZE] = (unsigned char*[DATA_SIZE]) &data;

// this gives a warning and doesn't work at all
unsigned char (* matrix)[DATA_SIZE] = (unsigned char **) &data;

我想将指向任意数据的指针转换为二维数组，但当然我无法转换为数组类型。我怎么需要这个呢？

提前致谢。

Answer 1

你得到了强制转换的语法，

unsigned char (* matrix)[DATA_SIZE] = (unsigned char(*)[DATA_SIZE]) data;

通过删除#define

修复=后，

正常工作

#define DATA_SIZE 10

    unsigned char *data = malloc(DATA_SIZE * DATA_SIZE * sizeof(unsigned char));

    // this is the correct way to cast
    unsigned char (* matrix)[DATA_SIZE] = (unsigned char(*)[DATA_SIZE]) data;

在C中使用char数组作为二维数组

1 个答案: