IA32 indexed addressing mode without base address

Question

What does this syntax mean? I'm specifically unsure about the missing base address

mov    0x804a1d4(,%edx,4),%eax

score 2 · Accepted Answer · edited Nov 11 '12 at 01:52

2

assign the contents of 0x804a1d4 + %edx*4 to %eax

edited Nov 11 '12 at 01:52

Community

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answered Nov 09 '11 at 04:54

steve

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That is backwards. The contents of the address (0x804a1d4 + edx * 4) are assigned to eax. – Borodin Nov 09 '11 at 07:10

score 1 · Answer 2 · answered Nov 09 '11 at 05:19

1

There is no base. Either the base register or the offset register (but not both) may be omitted. In this case the base address is hard-coded and the value to be moved into eax is loaded from

  0x804a1d4 + (edx * 4)

answered Nov 09 '11 at 05:19

Borodin

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IA32 indexed addressing mode without base address

2 Answers2