Machine level obfuscation
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# Machine level obfuscation

Let’s start with this small C program. What do you think it does?

1 2 3 4 5 6 7 #include <stdio.h> double d[]= {1156563417652693958656.0, 272}; int main() { d[1]--?d[0]*=2, main() : printf("%s\n",(char*)d); return 0; } 

Go ahead and run it on your machine. For lazy bums, you can see the output here. Well, the output of this code depends on the machine, more specifically the endianness of the machine. Let us walk through the code line by line to understand what is happening.

## Line 2

1 double d[]= {1156563417652693958656.0, 272}; 

Here we have simply declared a one-dimensional double array and initialized it with two elements with some values. The numbers are specific, which we shall see later in this post.

## Line 5

This line is a fancy way of saying

1 2 3 4 5 6 7 8 9 10 11 if(d[1] > 0) { d[1] = d[1] - 1; d[0] = 2 * d[0]; main(); } else { d[1] = d[1] - 1; printf("%s\n",(char*)d); } 

Ternary operators are used here instead of an if-else block to condense the code. The main function is called repeatedly until d[1] becomes 0. Then we typecast the double array to a char pointer and print its value as a string using the "%s" placeholder in the printf function.

So when does d[1] become 0? You got it right, it’s after doubling d[0] 272 times. Now when d[1] becomes 0, d[0] = 8.77663973968813359063877158122E102 in the mantissa exponent notation. This 64 bit double number in binary is represented as: 01010101 01001111 01011001 01000101 01010110 01001111 01001100 01001001

These are the 8 bytes (64 bits) representing the number. Here the first bit 0 denotes the sign (+ve) of the number. The next 11 bits 10101010100 represent the exponent, and the last 52 bits 1111 0101 1001 0100 0101 0101 0110 0100 1111 0100 1100 0100 1001 represent the mantissa.

This is an 8-byte representation (64 bits) in the IEEE754 standard and so a char pointer will read the whole thing one byte at a time, as char is 1-byte in size in the C language implementation. Now comes the role of endianness. On little endian machines, the string will be read backwards, i.e. the last byte is read first, so the bytes read in order are (binary, hex, decimal, char ascii):

• 01001001 = 0x49 = 73 = I
• 01001100 = 0x4C = 76 = L
• 01001111 = 0x4F = 79 = O
• 01010110 = 0x56 = 86 = V
• 01000101 = 0x45 = 69 = E
• 01011001 = 0x59 = 89 = Y
• 01001111 = 0x4F = 79 = O
• 01010101 = 0x55 = 85 = U

As d[] is an array, the next byte after d[0] in memory stores d[1] which is now = -1.0; so the last byte is 0000 which acts as a NULL character, terminating the string for the %s placeholder in printf. Hence, the output of the code above is ILOVEYOU.

Interesting, isn’t it? You just found a geeky way to say this to the love of your life! Anyway, this interesting aspect can be used to obfuscate any string into numbers, as I did with my name. ASEEMRAJ can be obfuscated with d[0] = 4875566432211777.0 and d[1] = 113. Now go and find the magical numbers for your own strings.

Note: If a float is used instead of double we have 4 bytes, hence 4 characters instead of 8.