文件加密算法求助

bikkuri

大家好！
我有一个问题向大家请教！

假如有两个二进制文件a和b，其中a为明文文件(文件长度为x字节)，b为密匙文件(文件长度为y字节)。

希望用以下算法用密匙文件对明文文件进行加密：

1. 明文文件a        运算        密钥文件b
   
2. 第一字节                XOR        最后一字节
   
3. 第二字节                XOR        倒数第二字节
   
4. 第n字节                XOR        第y-((n-1) mod y)字节 (其中0<n<y+1)
   
5. ......
   
6. 第y字节                XOR        第一字节
   
7. 第y+1字节        XOR        最后一字节
   
8. ...
   
9. 第x字节                XOR        第y-((x-1) mod y)字节

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希望大家帮忙用shell脚本实现这个加密算法。

1. cat a|file_xor b|file_xor b > c
   
2. diff a c

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明文文件a经过两次用同一个密钥文件加密后得到的结果应该和原来的明文文件相同。

谢谢大家！

jason680

回复 10# bikkuri

run only, like as below:

http://bbs.chinaunix.net/forum.p ... ;page=1#pid24467396
   

sync_1521

1. awk 'BEGIN{FS=OFS=""}NR==FNR{for(i=1;i<=NF;i++){a[++j]=$i};next}{for(i=1;i<=NF;i++){++k;$i=xor($i,a[k%j])}}1' b a

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bikkuri

谢谢您的回复！
不过显然加密处理后的结果是错误的。

1. [root@hp8:~/newtest]# date > a
   
2. [root@hp8:~/newtest]# date +%s > b
   
3. [root@hp8:~/newtest]# ls -l
   
4. total 8
   
5. -rw-r--r-- 1 root root 29 Jan  5 17:09 a
   
6. -rw-r--r-- 1 root root 11 Jan  5 17:09 b
   
7. [root@hp8:~/newtest]# awk 'BEGIN{FS=OFS=""}NR==FNR{for(i=1;i<=NF;i++){a[++j]=$i};next}{for(i=1;i<=NF;i++){++k;$i=xor($i,a[k%j])}}1' b a
   
8. 14519849851521914131101451118515
   
9. [root@hp8:~/newtest]# awk 'BEGIN{FS=OFS=""}NR==FNR{for(i=1;i<=NF;i++){a[++j]=$i};next}{for(i=1;i<=NF;i++){++k;$i=xor($i,a[k%j])}}1' b a > c
   
10. [root@hp8:~/newtest]# ls -l
    
11. total 12
    
12. -rw-r--r-- 1 root root 29 Jan  5 17:09 a
    
13. -rw-r--r-- 1 root root 11 Jan  5 17:09 b
    
14. -rw-r--r-- 1 root root 33 Jan  5 17:10 c
    
15. [root@hp8:~/newtest]# awk 'BEGIN{FS=OFS=""}NR==FNR{for(i=1;i<=NF;i++){a[++j]=$i};next}{for(i=1;i<=NF;i++){++k;$i=xor($i,a[k%j])}}1' b c > d
    
16. [root@hp8:~/newtest]# ls -l
    
17. total 16
    
18. -rw-r--r-- 1 root root 29 Jan  5 17:09 a
    
19. -rw-r--r-- 1 root root 11 Jan  5 17:09 b
    
20. -rw-r--r-- 1 root root 33 Jan  5 17:10 c
    
21. -rw-r--r-- 1 root root 35 Jan  5 17:10 d
    
22. [root@hp8:~/newtest]#
    

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回复 2# sync_1521


   

sync_1521

回复 3# bikkuri

1. test@server ~> cat a
   
2. 0101010101111111111
   
3. 0101010
   
4. 111
   
5. test@server ~> cat b
   
6. 010101
   
7. 10
   
8. test@server ~> awk 'BEGIN{FS=OFS=""}NR==FNR{for(i=1;i<=NF;i++){a[++j]=$i};next}{for(i=1;i<=NF;i++){++k;$i=xor($i,a[k%j])}}1' b a
   
9. 0000001100101001101
   
10. 1110011
    
11. 101
    
12. test@server ~> awk 'BEGIN{FS=OFS=""}NR==FNR{for(i=1;i<=NF;i++){a[++j]=$i};next}{for(i=1;i<=NF;i++){++k;$i=xor($i,a[k%j])}}1' b a >c
    
13. test@server ~> awk 'BEGIN{FS=OFS=""}NR==FNR{for(i=1;i<=NF;i++){a[++j]=$i};next}{for(i=1;i<=NF;i++){++k;$i=xor($i,a[k%j])}}1' b c >d
    
14. test@server ~> cat d
    
15. 0101010101111111111
    
16. 0101010
    
17. 111
    
18. test@server ~>

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要把文件搞成010101这种形式的吧？
   

jason680

回复 1# bikkuri

Note: debug by yourself with busybox and awk null (ASCII 0) issue

$ awk 'BEGIN{RS="@#*#@";for(n=1;n<=255;n++)a[sprintf("%c",n)]=n}FNR==NR{y=length-1;for(n=1;n<=y;n++)b[n]=a[substr($0,n,1)];next}{x=length;s="";for(n=1;n<=x;n++)s=s sprintf("%c",xor(a[substr($0,n,1)],b[y-n%y]));printf("%s",s)}' b a  > a.x

$ awk 'BEGIN{RS="@#*#@";for(n=1;n<=255;n++)a[sprintf("%c",n)]=n}FNR==NR{y=length-1;for(n=1;n<=y;n++)b[n]=a[substr($0,n,1)];next}{x=length;s="";for(n=1;n<=x;n++)s=s sprintf("%c",xor(a[substr($0,n,1)],b[y-n%y]));printf("%s",s)}' b a.x  > c

$ diff a c

$ ls -l a b c
-rw-rw-r-- 1 jason jason 29 Jan  5 17:45 a
-rw-rw-r-- 1 jason jason 11 Jan  5 17:44 b
-rw-rw-r-- 1 jason jason 29 Jan  5 18:07 c

$ cat a b c
Tue Jan  5 17:45:03 CST 2016
1451987093
Tue Jan  5 17:45:03 CST 2016

   

bikkuri

谢谢您的回复！
确实在Linux上您的程序可以得到正确的结果，但是在busybox上执行出错。
我研究一下看看在busybox上要怎么改。
另外您的程序中的RS="@#*#@"是什么意思啊？
好像从来没有见过这种写法，网上搜索了也没找到。
和RS=“”有什么区别吗？

回复 5# jason680


   

jason680

回复 6# bikkuri

https://www.gnu.org/software/gawk/manual/gawk.html#Multiple-Line

4.8 Multiple-Line Records
...

You can achieve the same effect as ‘RS = ""’ by assigning the string "\n\n+" to RS. This regexp matches the newline at the end of the record and one or more blank lines after the record. In addition, a regular expression always matches the longest possible sequence when there is a choice (see Leftmost Longest). So, the next record doesn’t start until the first nonblank line that follows—no matter how many blank lines appear in a row, they are considered one record separator.
...



https://www.gnu.org/software/gawk/manual/gawk.html#gawk-split-records

4.1.2 Record Splitting with gawk

                                           RS = "\0" Is Not Portable
There are times when you might want to treat an entire data file as a single record. The only way to make this happen is to give RS a value that you know doesn’t occur in the input file. This is hard to do in a general way, such that a program always works for arbitrary input files.

You might think that for text files, the NUL character, which consists of a character with all bits equal to zero, is a good value to use for RS in this case:

BEGIN { RS = "\0" }  # whole file becomes one record?

gawk in fact accepts this, and uses the NUL character for the record separator. This works for certain special files, such as /proc/environ on GNU/Linux systems, where the NUL character is in fact the record separator. However, this usage is not portable to most other awk implementations.

Almost all other awk implementations19 store strings internally as C-style strings. C strings use the NUL character as the string terminator. In effect, this means that ‘RS = "\0"’ is the same as ‘RS = ""’. (d.c.)

It happens that recent versions of mawk can use the NUL character as a record separator. However, this is a special case: mawk does not allow embedded NUL characters in strings. (This may change in a future version of mawk.)

See Readfile Function, for an interesting way to read whole files. If you are using gawk, see Extension Sample Readfile, for another option.
   

bikkuri

啊？原来RS="@#*#@"是用一个不可能出现的RS来把整个文件当作一个Record。
但是如果要处理的文件就是程序本身的话，这个RS就会把文件分成两个Record吧。
我觉得用RS=“$(md5sum $0|awk '{print$1}')”的话，也就是用文件的md5校验和来做RS，就可以确保RS不会出现在文件里吧？

回复 7# jason680


   

jason680

回复 8# bikkuri

>> ...但是如果要处理的文件就是程序本身的话...

说说你要做什么....
也可以只加密可打印字符(ASCII 32~126)
你说的我也曾做过...密码缺省则用档名