需要帮助读取二进制文件到结构 [英] Need help reading binary file to a structure

问题描述

我真的需要帮助这一个。我尝试下面的问题的解决方案，类似于我的，但输出是错误的。

I really need help on this one. I've tried following the solution of a question that was similar to mine, but the output is wrong.

文件是小端。我正在运行Win7，这也是小endian

The file is in little endian. I am running Win7, which is also little endian

我有一个struct Header，我想读取二进制文件到结构中。结构中的成员类型由问题定义。

I have a struct Header, and I want to read the binary file into the structure. The member types in the struct are defined by the problem.

struct Header
{
    uint16_t marker;
    uint8_t msg_type;
    uint64_t sequence_id;
    uint64_t timestamp;
    uint8_t msg_direction;
    uint16_t msg_len;
};
void parseFile()
{
  Header h;
  ifstream file("example_data_file.bin", std::ios::binary);

  file.read((char*)&h, sizeof(h));

  file.close();
  printf("%u \n",(int)h.marker);
  printf("%u \n",h.msg_type);
  printf("%u \n", h.sequence_id);
  printf("%u \n", h.timestamp);
  printf("%u \n",h.msg_direction);
  printf("%u \n",h.msg_len);
}

但是，输出没有意义。我预计第一行是ST，第二行是1,2或3.

However, the output is none sense. I am expecting the first row to be "ST", and the second row to be 1,2 or 3.

21587
1
1
3719304632
0
48

我也尝试过读这个struct的每个成员。

I have also tried reading each member of the struct individually like this

  file.read ((char*)&h.marker, sizeof(h.marker));
  file.read ((char*)&h.msg_type, sizeof(h.msg_type));
  file.read ((char*)&h.sequence_id, sizeof(h.sequence_id));
  file.read ((char*)&h.timestamp, sizeof(h.timestamp));
  file.read ((char*)&h.msg_direction, sizeof(h.msg_direction));
  file.read ((char*)&h.msg_len, sizeof(h.msg_len));

输出不同，但仍然错误。

The output is different, but it is still wrong.

以下是各个阅读的输出：

Here is the output of the individual reads:

21587
☺
1
1398642639377848

48

http://www.filedropper.com/exampledatafile

我不知道我做错了什么，我从来没有解析过二进制文件，所以这是我第一次做它。请帮助〜

I have no idea what im doing wrong, and I've never parsed a binary file before so this is my first time doing it. Please help~

谢谢！

推荐答案

打印数据的方式。很可能你的环境有

The error is in your way of printing the data. Likely, your environment has

typedef uint8_t char;

这是为什么您的 uint8_t 作为ASCII字符。类似地，ST是 0x53 0x54 这是小端头的21587。

which is why your uint8_ts get printed as an ASCII character. Similarly, "ST" is 0x53 0x54 which is 21587 in little endian.

使用以下命令打印出来：

Use following to print it out:

printf("%c%c\n", *(char *)&(h.marker), *((char *)&(h.marker))+1);
printf("%u\n", (unsigned int)h.msg_type);
printf("%llu\n", h.sequence_id);
printf("%llu\n", h.timestamp);
printf("%u\n", (unsigned int)h.msg_direction);
printf("%hu\n", h.msg_len);

*（char *）&（h.marker）的简短说明：

• 取地址 h.marker ： &（。h.marker）


• 解释为指向 char ：（char *）


• 取指针的内容： * （初始*）

• Take address of h.marker: &(.h.marker)
• Interpret it as a pointer to char: (char *)
• And take the pointer's content: * (the initial *)

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