以十六进制表示的4字节日期时间转换为可读

Question

I have a response from a server that was compressed and converted it to base64 string. I've decoded this response and found it to be something that looks like a soap document. Most values are in hex format. So I have a parameter datetime which stores value "\\xf4\\xdd|\\xad\\x08". I know that it is a representation of the 24.06.1982 date.

How I could convert hex value to the date format? Prefer c/c++ language.

Answer 1

To convert the encoded number to an integer, you need to decide if it's big-endian or little-endian.

Since the data you have is not a string, ie the number is not in text but directly in binary, you cannot sanely use sscanf() which is a string-parsing function.

Instead, so something like this:

const char *buffer = "\xf4\xdd|\xad\x08";
unsigned int stamp = (((unsigned char) buffer[0]) << 24) |
                     (((unsigned char) buffer[1]) << 16) |
                     (((unsigned char) buffer[3]) << 8) |
                     ((unsigned char) buffer[4]);

If it's little-endian, reverse the indexing sequence (so that buffer[0] becomes buffer[4] and so on).

Answer 2

to convert that hex format to a number, you can use scanf()

int a,b,c,d;
sscanf(input,"\\x%x\\x%x|\\x%x\\x%x",&a,&b,&c,&d);
long result=a<<24 | b<<16 | c<<8 | d;

That will give you 4108168456 for the value, you provided.

What further information about the format do you have?

int a,b,c,d;
sscanf(input,"\\x%x\\x%x|\\x%x\\x%x",&a,&b,&c,&d);
long field1=a<<8 | b;
long field2=c<<8 | d;