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> Main > i dont understand this..(the $ symbol) |
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badc0de02 01 May 2014, 19:46
$=current offset
jmp $ is the same as (loop forever) x: jmp x |
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01 May 2014, 19:46 |
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revolution 02 May 2014, 02:25
You can think of $ as always equalling {E|R}IP.
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02 May 2014, 02:25 |
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Bargest 02 May 2014, 14:58
revolution wrote: You can think of $ as always equalling {E|R}IP. Not EIP, but EIP minus size of current command. Because Code: jmp $ is equal to Code: label: jmp label But at this point EIP is equal to address of the next command. And this code will compile as 0xEB 0xFE, which means "jump -2". _________________ jmp $ ; Happy end! |
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02 May 2014, 14:58 |
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revolution 02 May 2014, 15:35
Sure. But that is merely an implementation issue of the x86. Other CPUs can do different things. At the assembly level $ is the same as IP in the logical sense of being the address of the current instruction.
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02 May 2014, 15:35 |
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shutdownall 03 May 2014, 17:00
revolution wrote: You can think of $ as always equalling {E|R}IP. Well - depends on type of segment. If you create a data segment (no execute) it wouldn't be suitable to compare it with the actual position of IP. Or when using segmented addresses in real mode. |
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03 May 2014, 17:00 |
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badc0de02 04 May 2014, 08:24
$=current offset
$$=offset from the begin of you code Code: org 0x7c00 boot: jmp boot ; loop times 512-($-$$) db 0 ; times 512-((0x7c00+2)-0x7c00) |
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04 May 2014, 08:24 |
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