Hey there! Let's break down this x86-64 assembly instruction clearly since you're new to assembly—this addressing mode is really common once you wrap your head around it.

First, let's recap what the entire instruction does: it copies the 64-bit value from the rdx register into a memory location defined by the expression in parentheses. The part (%1, %%r8, 8) uses x86-64's base + index + scale addressing mode, which follows this structure:

(base_register, index_register, scale_factor)

Let's break down each component, with a focus on that 8:

• %1: This is a placeholder (typically from GCC-style inline assembly) that represents a base register (like rax, rbx, etc.) holding a starting memory address.
• %%r8: This is the index register—its value will be scaled (multiplied) by the factor we're talking about.
• 8: This is the scale factor. Its job is to multiply the value in the index register (r8) to calculate the offset from the base address.

Why do we need this scale factor? #

x86-64 uses this to efficiently access arrays or structs where each element is a fixed size. Since you're using movq (a 64-bit move instruction), you're likely dealing with 8-byte elements (like 64-bit integers or pointers). The scale factor lets the CPU automatically compute the correct offset without you having to write an explicit multiply instruction.

The formula for the final memory address is:

Final Address = [Value of %1] + ([Value of %r8] × 8)

Example to make it concrete:

If %1 holds the address 0x1000 (the start of an array of 8-byte values) and %r8 holds 3 (the index of the element we want), the calculation would be:
0x1000 + (3 × 8) = 0x1018
The movq instruction would then write the value from rdx into the memory at 0x1018—which is exactly the 4th element in our 8-byte array (since indexes start at 0).

Note that valid scale factors in x86-64 are 1, 2, 4, and 8—each corresponding to common data sizes (byte, 16-bit, 32-bit, 64-bit) to make array accesses straightforward.

内容的提问来源于stack exchange，提问作者assembly_question