将二进制数据（从文件）读取到结构中 [英] Read binary data (from file) into a struct
问题描述
我正在从文件（尤其是从zip文件）读取二进制数据。 （要了解有关zip格式结构的更多信息，请参见 http://en.wikipedia.org/wiki / ZIP_％28file_format％29 ）
I'm reading binary data from a file, specifically from a zip file. (To know more about the zip format structure see http://en.wikipedia.org/wiki/ZIP_%28file_format%29)
我创建了一个存储数据的结构：
I've created a struct that stores the data:
typedef struct {
/*Start Size Description */
int signatute; /* 0 4 Local file header signature = 0x04034b50 */
short int version; /* 4 2 Version needed to extract (minimum) */
short int bit_flag; /* 6 2 General purpose bit flag */
short int compression_method; /* 8 2 Compression method */
short int time; /* 10 2 File last modification time */
short int date; /* 12 2 File last modification date */
int crc; /* 14 4 CRC-32 */
int compressed_size; /* 18 4 Compressed size */
int uncompressed_size; /* 22 4 Uncompressed size */
short int name_length; /* 26 2 File name length (n) */
short int extra_field_length; /* 28 2 Extra field length (m) */
char *name; /* 30 n File name */
char *extra_field; /*30+n m Extra field */
} ZIP_local_file_header;
sizeof（ZIP_local_file_header）返回的大小是40，但是如果使用 sizeof 运算符计算每个字段的总和，则总大小为38。
The size returned by sizeof(ZIP_local_file_header) is 40, but if the sum of each field is calculated with sizeof operator the total size is 38.
如果我们有下一个结构：
If we have the next struct:
typedef struct {
short int x;
int y;
} FOO;
sizeof（FOO）返回8，因为每次分配4个字节的内存。因此，要分配 x 保留4个字节（但实际大小为2个字节）。如果我们需要另一个 short int ，它将填充先前分配的其余2个字节。但是，由于我们有一个 int ，它将被分配加上4个字节，并且浪费了2个字节。
sizeof(FOO) returns 8 because the memory is allocated with 4 bytes every time. So, to allocate x are reserved 4 bytes (but the real size is 2 bytes). If we need another short int it will fill the resting 2 bytes of the previous allocation. But as we have an int it will be allocated plus 4 bytes and the empty 2 bytes are wasted.
To从文件中读取数据，我们可以使用函数 fread ：
To read data from file, we can use the function fread:
ZIP_local_file_header p;
fread(&p,sizeof(ZIP_local_file_header),1,file);
但是由于中间有空字节，因此无法正确读取。
But as there're empty bytes in the middle, it isn't read correctly.
如何使用 ZIP_local_file_header 不浪费字节来顺序有效地存储数据？
What can I do to sequentially and efficiently store data with ZIP_local_file_header wasting no bytes?
推荐答案
C struct s只是将相关数据分组在一起，它们不指定内存中的特定布局。 （就像未定义 int 的宽度一样。）Little-endian / Big-endian也未定义，并且取决于处理器。
C structs are just about grouping related pieces of data together, they do not specify a particular layout in memory. (Just as the width of an int isn't defined either.) Little-endian/Big-endian is also not defined, and depends on the processor.
不同的编译器，在不同体系结构或操作系统上的同一编译器等，所有布局结构都将不同。
Different compilers, the same compiler on different architectures or operating systems, etc., will all layout structs differently.
要读取的文件格式是根据字节的位置来定义的，尽管结构看起来非常方便和诱人，但它不是正确的解决方案。您需要将文件视为 char [] 并取出所需的字节并进行移位，以使数字由多个字节等组成。
As the file format you want to read is defined in terms of which bytes go where, a struct, although it looks very convenient and tempting, isn't the right solution. You need to treat the file as a char[] and pull out the bytes you need and shift them in order to make numbers composed of multiple bytes, etc.
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