CODEWITHSUNDEEP
disassembly
mithya
mithya

Reputation: 363

What does comma separated bracket mov instruction mean in disassembly

   0x7f52379dc42c:  mov    0xc(%r12,%r11,8),%r11d
   0x7f52379dc431:  mov    %r11d,0xc(%rsp)
   0x7f52379dc436:  mov    0xc(%r12,%r10,8),%r14d
   0x7f52379dc43b:  cmp    %r11d,%r14d

I understand that mov %r11d,0xc(%rsp) means *(rsp+0xc) = 0xc

What does mov 0xc(%r12,%r11,8),%r11d mean?

Upvotes: 2

Views: 770

Answers (1)

Erlend Graff
Erlend Graff

Reputation: 1548

The general syntax for a memory operand (dereference) in AT&T x86/x64 mnemonics is offset(base, index, scale), which is the same as [base + index * scale + offset] in Intel syntax (which is almost the same as the pseudo-C syntax you used).

Specifically, your first instruction

mov 0xc(%r12,%r11,8), %r11d

is the same as

mov r11d, DWORD PTR [r12+r11*8+0xc]

in Intel mnemonics, and approximately the same as

r11d = *(r12 + r11 * 8 + 0xc)

in the pseudo-C syntax.

Note that the scale is encoded using only 2 bits in the instruction, and is always a power-of-two, so only values of 1, 2, 4, and 8 are permitted.

Upvotes: 2

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