Simply, why does not call 12345678 jump to 12345678 address?
Why do I have to use the instruction like this
mov eax, 12345678
call eax
More importantly, what does call 12345678 exactly do?
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2 Answers
Because of call binary encoding. See the details and properties for intel processors at
Intel manual at page 3-114 Vol. 2A for Intel architecture. If you are talking about different processor, please state which exactly.
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In the latest Intel Manual this page (3-114 Vol. 2A) refers to the
CLIinstruction documentation, however just a few pages back (3-96 Vol. 2A; Page 575-580) contains a complete treatment ofCALLbehavior.– zetavoltJul 30, 2015 at 20:26
call 12345678 pushes the following instruction's return address onto the stack and then jumps to 12345678.
It is functionally equivalent to:
mov eax, 12345678
call eax
(Though the code above has the side-effect of modifying eax, whereas a simple call 12345678 does not modify eax.)
answered Jul 27, 2015 at 13:13
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All variants of call with immediate operands are relative, which may lead to the problem described in the question. The only exception is 9a opcode which is not valid in 64 bit mode.– w sJul 27, 2015 at 14:14
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Yes, but @FreeMind was asking about the instruction (the "disassembly"), not the machine code. Jul 27, 2015 at 14:41