我知道转换为十进制时的二进制值“1000001”是65，根据ASCII表是字母A.
现在，我有一个二进制文件（xxx.bin），假设有上面的位序列。我怎么知道tat二进制值是表示十进制值65还是字符A ??
我想当我收到二进制文件时，我应该也知道这个二进制文件遵循什么字符编码或字符集doc？或者我能以某种方式找出它是如何被编码的吗？
有人可以清楚如何正确解码二进制文件（xxx.bin）数据并读取？只需将每个字节转换为十进制值或遵循一些字符编码逻辑???我对此感到困惑!!
答案 0 :(得分：0)
这是我将字符串编码为'dots'和'spaces'的程序。如果提供代码，它还可以将其解码回原始字符串。亲自尝试一下。其中有一个encode_find（）函数，它将一个字符转换为二进制等价物并将其存储在变量“en”中。反向过程可以实现decode_find（）函数。
import java.io.*;
class ed2
{
BufferedReader obj = new BufferedReader(new InputStreamReader(System.in));
void encode()throws InterruptedException,IOException
{
String rs,ren;
encoder_symbol();
rs = encode_input();
ren = encode_find(rs);
encode_display(ren);
}
void decode()throws InterruptedException,IOException
{
String rs,rde;
decoder_symbol();
rs = decode_input();
rde = decode_find(rs);
decode_display(rde);
}
void encoder_symbol()throws InterruptedException //just for fun
{
System.out.println("******** *** *** ********* ************ ****** ******** *****");
Thread.sleep(100);
System.out.println("******** **** *** ********* ************ ******** ******** *** **");
Thread.sleep(100);
System.out.println("*** ***** *** *** *** *** *** *** *** *** **");
Thread.sleep(100);
System.out.println("*** *** ** *** *** *** *** *** *** *** *** **");
Thread.sleep(100);
System.out.println("****** *** ** *** *** *** *** *** *** ****** *****");
Thread.sleep(100);
System.out.println("****** *** ** *** *** *** *** *** *** ****** *****");
Thread.sleep(100);
System.out.println("*** *** ** *** *** *** *** *** *** *** *** **");
Thread.sleep(100);
System.out.println("*** *** ** *** *** *** *** *** *** *** *** **");
Thread.sleep(100);
System.out.println("******* *** ***** ********* ************ ******** ******** *** **");
Thread.sleep(100);
System.out.println("******* *** **** ********* ************ ****** ******** *** **");
Thread.sleep(2700);
System.out.println();
System.out.println();
}
void decoder_symbol()throws InterruptedException // just for fun
{
System.out.println("****** ******** ********* ************ ****** ******** *****");
Thread.sleep(100);
System.out.println("******** ******** ********* ************ ******** ******** *** **");
Thread.sleep(100);
System.out.println("*** *** *** *** *** *** *** *** *** *** **");
Thread.sleep(100);
System.out.println("*** *** *** *** *** *** *** *** *** *** **");
Thread.sleep(100);
System.out.println("*** *** ****** *** *** *** *** *** ****** *****");
Thread.sleep(100);
System.out.println("*** *** ****** *** *** *** *** *** ****** *****");
Thread.sleep(100);
System.out.println("*** *** *** *** *** *** *** *** *** *** **");
Thread.sleep(100);
System.out.println("*** *** *** *** *** *** *** *** *** *** **");
Thread.sleep(100);
System.out.println("******** ******** ********* ************ ******** ******** *** **");
Thread.sleep(100);
System.out.println("****** ******** ********* ************ ****** ******** *** **");
Thread.sleep(1000);
System.out.println();
System.out.println();
}
String encode_input()throws IOException
{
String s;
System.out.println("ENTER THE STRING TO BE ENCODED");
s = obj.readLine();
return(s);
}
String decode_input()throws IOException
{
String s;
System.out.println("ENTER THE CODE TO BE DECODED");
s = obj.readLine();
return(s);
}
String encode_find(String s)//converting the string into its binary equivalent
{
int ac,i,j,l,chklen;
String bc,en="";
char ic;
l = s.length();
for(i=0;i<l;i++)
{
ic = s.charAt(i); //takes out every character
bc = "";
ac = (int)ic; //ASCII value of this character
while(ac!=0)
{
bc = Integer.toString((ac%2)) + bc; //converting the ASCII value into binary equivalent
ac = ac/2;
}
chklen = bc.length();//length of the binary equivalent
if(chklen<7)
{
for(j=1;j<=(7-chklen);j++) //increasing the length of binary equivalent so that it becomes equal to 7
{
bc = "0" + bc;
}
}
en = en+bc; //concatenating all the binary equivalent into one string
}
return (en);
}
String decode_find(String s)// converts binary(i.e. in the form of dots and space) to decimal
{
int f;//for the index of every character of code
long l,i,j,ac;
char c;
String de="";
l = s.length();
f = 0;//index of first caharcter
for(i=0;i<(l/7);i++)//since the length of every binary equivalent of a character is 7 therefore there will be (length/7) characters in a code of length l
{
ac = 0;//intializes the decimal(ASCII) equivalent to zero
for(j=6;j>=0;j--)//loop will work 7 times for every binary equivalent of a character
{
c = s.charAt(f);//takes out every dot or space
if(c=='.')//it means that c corresponds to 'one'
{
ac = ac + ((int)Math.pow(2,j));//converting binary into decimal(ASCII) equivalent by adding all the powers of 2 which correspond to one('.')
}
f++;//increasing the index for next character of binary equivalent
}
de = de + ((char)ac);//converts the ASCII equivalent into character and then concatenates it with the intitial string
}
return(de);
}
void encode_display(String en)//displays the code
{
int i,l;
char ic;
System.out.println("YOUR ENCODED MESSAGE IS :");
l=en.length();
for(i=0;i<l;i++)
{
ic=en.charAt(i);
if(ic=='1')//for every 'one' it will print '.'(dot)
{
System.out.print(".");
}
else if(ic=='0')//for every 'zero' it will print ' '(space)
{
System.out.print(" ");
}
}
}
void decode_display(String de)
{
System.out.println(de);
}
public static void main(String args[])throws IOException,InterruptedException
{
BufferedReader obj = new BufferedReader(new InputStreamReader(System.in));
char ch;
ed2 ed = new ed2();
System.out.println("PRESS 'E' TO ENCODE A MESSAGE OR PRESS 'D' TO DECODE A MESSAGE");
ch = (char)obj.read();
if((ch=='e')||(ch=='E'))
{
ed.encode();
}
else if((ch=='d')||(ch=='D'))
{
ed.decode();
}
}
}
答案 1 :(得分：0)
如果您不知道文件的格式，则会丢失数据。
问题是如何防止错过重要的“元数据”。典型的方法是通过协议和规范。
在某些情况下，您可以确定文件具有在某些编码中无效的字节值或序列，例如0x80无效ASCII和0xE2 0x20无效UTF-8。有这种类型的猜测库。但是，永远不能排除CP437;因此，每个文件都可以被视为文本文件，就像每个文件可以被认为是0-255或-128到127的值序列一样，并且可以考虑具有偶数字节数的每个文件....
有协议分析器根据各种嵌套协议解释字节，例如Wireshark，还有Hex编辑器显示字节和尝试转换为字符。一些十六进制编辑器非常复杂，具有按字节查找，按文本查找，允许您描述协议并将其应用于文件的某些部分等功能。 （旧的磁盘扇区编辑器就是一个例子。）
最重要的是，这是逆向工程，希望你不需要这样做。
BTW-65是许多编码中的字母A;不只是ASCII。