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*(_BYTE *)(Key + Integer_count - 1) + 79

Here : Key = 0x22EC278 (dw 'Âx') or

Stack[000015B8]:0019FC2C db 78h ; x

Stack[000015B8]:0019FC2D db 0C2h ; Â

Stack[000015B8]:0019FC2E db 2Eh ; .

Stack[000015B8]:0019FC2F db 2

and integer_count is just a incremented loop (e.g. 1,2,3,4,...)

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  • Key is a pointer it adds 79 to each byte upto integer count. Assume if key to key + intcount contains 0 then after this loop key to key+int count will all be 79 memset (* , 79 , intcount);
    – blabb
    Commented Mar 4, 2023 at 4:45
  • I am still little confused.... as this is entire function {String_from_char_((char *)&Decoded_string, *(_BYTE *)(Addressforencoded + Integer_count - 1) + 79);} now as you said {*(_BYTE *)(Addressforencoded + Integer_count - 1)} --> 0xFE and then addition happen so 0xFE+0x4f(hexof 79) --> gives 0x14D but only 4D has been written in memory
    – Satyam Singh
    Commented Mar 4, 2023 at 10:45
  • Yes correct overflow will be discarded byte is 8 bit max 0xff min 0x00. Ox14d is word or 16 bit. So obviously 0x100 is discarded
    – blabb
    Commented Mar 4, 2023 at 11:58
  • Ok Thanks Got it.
    – Satyam Singh
    Commented Mar 4, 2023 at 14:53

1 Answer 1

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-1

That expression is dereferencing a pointer (Key + Integer_count - 1) and then adding 79 to the dereferenced value. So for example, if:

Key = 0x22EC278 Integer_count = 3

Then it would evaluate to: *(0x22EC27B) + 79

Which is accessing the byte at address 0x22EC27B (the last byte of the Key buffer) and adding 79 to it. So in your specific examples, it would be: *(0x22EC2F) + 79 // accessing the last byte (0x2F) of Key and adding 79 *(0x22EC2E) + 79 // accessing the byte before the last (0x2E) and adding 79

etc. So it's basically incrementing the last byte(s) of the Key buffer by 79 on each iteration of the Integer_count loop.

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