您好,我的任务是将整个结构复制到新结构中,并分配结构内存。所以我想将一种结构复制到另一种结构中,也想复制它的内存。因此,如果我“释放”第一个结构,那么我将在内存中找到此信息。
struct bmp_image* flip_vertically(const struct bmp_image* image) {
struct bmp_image* bmp = NULL;
bmp = (struct bmp_image*)realloc(bmp, sizeof(struct bmp_image));
memcpy(bmp, image, sizeof(struct bmp_image));
return bmp;
}
可能是什么问题?
例如:
如果我知道
image-> data [5] .blue是255,我想复制它,但我还需要将该信息保存到内存中]
主要任务是翻转bmp图片。一切对我来说都很完美,但是如果我把这个源代码进行测试,则写成:
Running suite(s): flip_horizontally()
stderr
double free or corruption (!prev)
这意味着测试在某个地方释放了旧结构,因此我在新的结构中没有此信息
struct bmp_header{
uint16_t type; // "BM" (0x42, 0x4D)
uint32_t size; // file size
uint16_t reserved1; // not used (0)
uint16_t reserved2; // not used (0)
uint32_t offset; // offset to image data (54B)
uint32_t dib_size; // DIB header size (40B)
uint32_t width; // width in pixels
uint32_t height; // height in pixels
uint16_t planes; // 1
uint16_t bpp; // bits per pixel (1/4/8/24)
uint32_t compression; // compression type (0/1/2) 0
uint32_t image_size; // size of picture in bytes, 0
uint32_t x_ppm; // X Pixels per meter (0)
uint32_t y_ppm; // X Pixels per meter (0)
uint32_t num_colors; // number of colors (0)
uint32_t important_colors; // important colors (0)
} __attribute__((__packed__));
/**
* This structure describes a color consisting of relative intensities of
* red, green, and blue.
*/
struct pixel {
uint8_t blue;
uint8_t green;
uint8_t red;
//uint8_t alpha;
} __attribute__((__packed__));
/**
* Structure describes the BMP file format, which consists from two parts:
* 1. the header (metadata)
* 2. the data (pixels)
*/
struct bmp_image {
struct bmp_header* header;
struct pixel* data; // nr. of pixels is `width` * `height`
};
Main.c]
int main () {
struct bmp_image* image = NULL;
image = read_bmp(stream);
FILE *output_p1 = fopen("square2.bmp", "w");
struct bmp_image* newimage1 = NULL;
newimage1 = flip_vertically(image);
free_bmp_image(image);
write_bmp(output_p1, newimage1);
free(newimage1);
fclose(output_p1);
return 0;
}
如果我释放图像(旧结构),它将为我显示很多错误,并且我无法将其写入文件。对我来说,这意味着它想从旧结构中读取。
memcpy
执行“浅”复制而不是“深”复制。浅表副本将仅复制struct bmp_image
中的指针值,而不复制其指向的内存。要进行深层复制,需要分配和复制各个字段。这是一些说明性代码。为简洁起见,未进行错误检查,但对于最终代码,应检查所有分配结果。
struct bmp_image* flip_vertically(const struct bmp_image* image) {
bmp = malloc(sizeof(*bmp));
bmp->header = malloc(sizeof(*(bmp->header)));
*bmp->header = *(image->header);
size_t pixel_data_size =
sizeof(*(bmp->data)) * bmp->header->width * bmp->header->height;
bmp->data = malloc(pixel_data_size);
memcpy(bmp->data, image->data, pixel_data_size);
return bmp;
}