fread（）中的空字符和c中的strncpy（）(Null character in fread() and strncpy() in c)

 有两种正确的方法来处理标题。 我假设MP3文件有IDV3标签，因此文件以“TAG”或“TAG +”开头。 因此，您要读取的部分有4个字节。 
 
 a）您认为char *memory是C“string”，first.header_id也是如此。 然后这样做（省略其他所有内容以显示重要部分）： 
 
typedef struct{
  unsigned char header_id[5];
} mp3_Header;
char memory[5];

fread(memory, 4, 1, file);
memory[4]='\0';
strncpy(first.header_id, memory, 5)
 
 在恐惧之后，你的记忆看起来像这样： 
 
   0    1    2    3    4
+----+----+----+----+----+
|  T |  A |  G |  + |  ? |
+----+----+----+----+----+
 
 索引4处的第5个字节未定义，因为您只读取4个字节。 如果对此字符串使用字符串函数（例如printf("%s\n", memory) ）; 该函数不知道在哪里停止，因为没有终止\ 0，并且printf将继续输出垃圾直到下一个\ 0它找到计算机RAM中的某个位置。 这就是为什么你接下来做memory[4]='\0'所以它看起来像这样： 
 
   0    1    2    3    4
+----+----+----+----+----+
|  T |  A |  G |  + | \0 |
+----+----+----+----+----+
 
 现在，您可以使用strncpy将这5个字节复制到first.header_id。 请注意，您需要复制5个字节，而不仅仅是4个字节，您还需要复制\ 0。 
 
 （在这种情况下，你也可以使用strcpy（没有n） - 它会在它遇到的第一个\ 0处停止。但是现在，为了防止缓冲区溢出，人们似乎同意不使用strcpy;相反，总是使用strncpy并明确说明接收字符串的长度）。 
 
 b）您将memory视为二进制数据，将二进制数据复制到标头， 然后将二进制数据转换为字符串： 
 
typedef struct{
  unsigned char header_id[5];
} mp3_Header;
char memory[4];

fread(memory, 4, 1, file);
memcpy(first.header_id, memory, 4)
first.header_id[4]='\0';
 
 在这种情况下，内存的末尾永远不会有\ 0。 所以现在使用4字节数组就足够了。 在这种情况下（复制二进制数据），你不使用strcpy，而是使用memcpy。 这只复制了4个字节。 但是现在， first.header_id没有结束标记，因此您必须明确指定它。 如果不是100％清楚的话，尝试像我上面那样绘制图像。 
 
 但请记住：如果您使用像'+'这样的运算符，则不会对字符串起作用。 你处理单个字符。 C语言中唯一能够处理字符串的方法是使用str *函数。 

There are 2 correct ways of handling the header. I'm assuming the MP3 file has a IDV3 tag, so the file starts with "TAG" or "TAG+". So the part you want to read has 4 bytes.
 
a) You think of char *memory being a C "string", and first.header_id as well. Then do it this way (omitted everything else to show the important parts):
 
typedef struct{
  unsigned char header_id[5];
} mp3_Header;
char memory[5];

fread(memory, 4, 1, file);
memory[4]='\0';
strncpy(first.header_id, memory, 5)
 
After the fread, your memory looks like this:
 
   0    1    2    3    4
+----+----+----+----+----+
|  T |  A |  G |  + |  ? |
+----+----+----+----+----+
 
The 5th byte, at index 4, is not defined, because you read only 4 bytes. If you use a string function on this string (for example printf("%s\n", memory)); the function doesn't know where to stop, because there is no terminating \0, and printf will continue to output garbage until the next \0 it finds somewhere in your computer's RAM. That's why you do memory[4]='\0' next so it looks like this:
 
   0    1    2    3    4
+----+----+----+----+----+
|  T |  A |  G |  + | \0 |
+----+----+----+----+----+
 
Now, you can use strncpy to copy these 5 bytes to first.header_id. Note you need to copy 5 bytes, not just 4, you want the \0 copied as well.
 
(In this case, you could use strcpy (without n) as well - it stops at the first \0 it encounters. But these days, to prevent buffer overflows, people seem to agree on not using strcpy at all; instead, always use strncpy and explicitly state the length of the receiving string).
 
b) You treat memory as binary data, copy the binary data to the header, and then turn the binary data into a string:
 
typedef struct{
  unsigned char header_id[5];
} mp3_Header;
char memory[4];

fread(memory, 4, 1, file);
memcpy(first.header_id, memory, 4)
first.header_id[4]='\0';
 
In this case, there is never a \0 at the end of memory. So it's sufficient to use a 4-byte-array now. In this case (copying binary data), you don't use strcpy, you use memcpy instead. This copies just the 4 bytes. But now, first.header_id has no end marker, so you have to assign it explicitly. Try drawing images like i did above if it isn't 100% clear to you.
 
But always remember: if you use operators like '+', you do NOT work upon the string. You work on the single characters. The only way, in C, to work on a string as a whole, is using the str* functions.