Thread: decoding

your decrypt function uses strlen(text) (each loop iteration - but it is not the point)

When string contains q character encrypted string will have 'q' ^ 'q' ==> 0 at this point

So next call to decrypt will not go further and stop decoding on this point.

You need to learn to use debugger

For example - do not encrypt/decrypt 'q' character

Your problem, as vart said, is that 'q' ^ 'q' == '\0' and '\0' is interpreted by the string functions as the end of the string.

You have three choices to fix it:
1. Treat 'q' as a special case, possibly leaving it as 'q'.
2. Instead of 'q', use a bit-pattern that won't occur in the input.
3. Use the "mem" functions instead of the "str" functions.

Originally Posted by vart

your decrypt function uses strlen(text) (each loop iteration - but it is not the point)

But the repeated calls to strlen will be optimized out when optimizations are active, so it's not such a terrible thing.

Originally Posted by oogabooga

But the repeated calls to strlen will be optimized out when optimizations are active, so it's not such a terrible thing.

Except when the length of the string could be changed in the loop as in this case

Originally Posted by vart

Except when the length of the string could be changed in the loop as in this case

Good point. Dumb of me not to see that. Here's the loop with "gcc -O3" and it's using "repnz scas" to find the end of the string again and again within the loop, as you said. So it has inlined and optimized strlen, but of course hasn't moved it out of the loop.

Code:

401348: 80 34 16 71          xorb   $0x71,(%esi,%edx,1)
40134c: 42                   inc    %edx
40134d: 89 e9                mov    %ebp,%ecx
40134f: 89 f7                mov    %esi,%edi
401351: f2 ae                repnz scas %es:(%edi),%al
401353: f7 d1                not    %ecx
401355: 8d 59 ff             lea    -0x1(%ecx),%ebx
401358: 39 da                cmp    %ebx,%edx
40135a: 72 ec                jb     401348 <_decrypt+0x14>

It's interesting how much more "optimization" gcc will do if you take the strlen() call out of the loop. Optimization is an arcane art.

Code:

void decrypt(char *text) {
     char key = 'q';
     size_t i;
     size_t len = strlen(text);
     for (i = 0; i < len; i++)
           text[i] ^= key;
}

00401334 <_decrypt>:
  401334:    55                       push   %ebp
  401335:    57                       push   %edi
  401336:    56                       push   %esi
  401337:    53                       push   %ebx
  401338:    52                       push   %edx

  401339:    8b 5c 24 18              mov    0x18(%esp),%ebx
  40133d:    31 c0                    xor    %eax,%eax
  40133f:    b9 ff ff ff ff           mov    $0xffffffff,%ecx
  401344:    89 df                    mov    %ebx,%edi

  401346:    f2 ae                    repnz scas %es:(%edi),%al
  401348:    f7 d1                    not    %ecx
  40134a:    49                       dec    %ecx
  40134b:    74 58                    je     4013a5 <_decrypt+0x71>

  40134d:    89 d8                    mov    %ebx,%eax
  40134f:    f7 d8                    neg    %eax
  401351:    83 e0 03                 and    $0x3,%eax
  401354:    89 ce                    mov    %ecx,%esi
  401356:    39 c1                    cmp    %eax,%ecx
  401358:    77 52                    ja     4013ac <_decrypt+0x78>
  40135a:    83 f9 04                 cmp    $0x4,%ecx
  40135d:    77 51                    ja     4013b0 <_decrypt+0x7c>
  40135f:    89 ce                    mov    %ecx,%esi
  401361:    31 c0                    xor    %eax,%eax
  401363:    90                       nop

  401364:    80 34 03 71              xorb   $0x71,(%ebx,%eax,1)
  401368:    40                       inc    %eax
  401369:    39 f0                    cmp    %esi,%eax
  40136b:    72 f7                    jb     401364 <_decrypt+0x30>

  40136d:    39 f1                    cmp    %esi,%ecx
  40136f:    74 34                    je     4013a5 <_decrypt+0x71>
  401371:    89 cd                    mov    %ecx,%ebp
  401373:    29 f5                    sub    %esi,%ebp
  401375:    89 ef                    mov    %ebp,%edi
  401377:    c1 ef 02                 shr    $0x2,%edi
  40137a:    89 fa                    mov    %edi,%edx
  40137c:    c1 e2 02                 shl    $0x2,%edx
  40137f:    89 14 24                 mov    %edx,(%esp)
  401382:    74 18                    je     40139c <_decrypt+0x68>

  401384:    01 de                    add    %ebx,%esi
  401386:    31 d2                    xor    %edx,%edx

  401388:    81 34 96 71 71 71 71     xorl   $0x71717171,(%esi,%edx,4)
  40138f:    42                       inc    %edx
  401390:    39 d7                    cmp    %edx,%edi
  401392:    77 f4                    ja     401388 <_decrypt+0x54>

  401394:    03 04 24                 add    (%esp),%eax
  401397:    3b 2c 24                 cmp    (%esp),%ebp
  40139a:    74 09                    je     4013a5 <_decrypt+0x71>

  40139c:    80 34 03 71              xorb   $0x71,(%ebx,%eax,1)
  4013a0:    40                       inc    %eax
  4013a1:    39 c1                    cmp    %eax,%ecx
  4013a3:    77 f7                    ja     40139c <_decrypt+0x68>

  4013a5:    58                       pop    %eax
  4013a6:    5b                       pop    %ebx
  4013a7:    5e                       pop    %esi
  4013a8:    5f                       pop    %edi
  4013a9:    5d                       pop    %ebp
  4013aa:    c3                       ret    

  4013ac:    89 c6                    mov    %eax,%esi
  4013ae:    eb aa                    jmp    40135a <_decrypt+0x26>
  4013b0:    85 f6                    test   %esi,%esi
  4013b2:    75 ad                    jne    401361 <_decrypt+0x2d>
  4013b4:    31 c0                    xor    %eax,%eax
  4013b6:    eb b9                    jmp    401371 <_decrypt+0x3d>

Thanks guys. But I still don't understand the strlen thing, can some one please explain?
And one more thing; if q ^ q = null ; then how come the second encryption and decryption worked?

/*second encryption and decryption*/ int ch;
while((ch = fgetc(fptr)) != EOF)
{
ch = ch ^ 'q'; //encrypt
ch = ch ^ 'q'; //decrypt
printf("%c",ch);
}

Originally Posted by Harith

Thanks guys. But I still don't understand the strlen thing, can some one please explain?
And one more thing; if q ^ q = null ; then how come the second encryption and decryption worked?

Putting the strlen() call in the test portion of the loop causes it to be executed on every iteration (with no chance of optimizing it out since its parameter is modified in the loop body). If instead you put it before the loop, storing the result in a variable that you use in the test portion of the loop, then it will only be executed once.


As for 'q' ^ 'q' being 0, the problem is that this inserts a 0 into your string. Since 0 indicates the end of the string, it effectively truncates your string at that point.


It works in the quoted portion above because you are not storing the result in a string this time, but simply applying the xor twice in a row to a single character and printing the result, which is the original ch.

when you'll need to store the encoded result - and later decode it - you may think about base64 encoding of the "encoded" string.