A couple of questions about [base + index*scale + disp]

The general form for memory addressing in Intel and AT&T Syntax is the following:

[base + index*scale + disp]
disp(base, index, scale)

My questions are the following:

• Can base and index be any register?
• What values can scale take, is it 1, 2, 4 and 8 (with 1 being the default)?
• Are index and disp interchangeable (with the only difference being that index is a register while disp is an immediate value)?

Solution 1:

This is described in Intel's manual:

3.7.5 Specifying an Offset
The offset part of a memory address can be specified directly as a static value (called a displacement) or through an address computation made up of one or more of the following components:

• Displacement — An 8-, 16-, or 32-bit value.
• Base — The value in a general-purpose register.
• Index — The value in a general-purpose register. [can't be ESP/RSP]
• Scale factor — A value of 2, 4, or 8 that is multiplied by the index value.

The offset which results from adding these components is called an effective address.

The scale-factor is encoded as a 2-bit shift count (0,1,2,3), for scale factors of 1, 2, 4, or 8. And yes, *1 (shift count = 0) is the default if you write (%edi, %edx); that's equivalent to (%edi, %edx, 1)

In AT&T syntax, it's disp(base, index, scale) - constants go outside the parens. Some Intel-syntax assemblers also allow syntax like 1234[ebx], others don't. But AT&T syntax is rigid; every component of the addressing mode can only go in its proper place. For example:

movzwl  foo-0x10(,%edx,2), %eax

does a zero-extending 16-bit ("word") load into EAX, from the address foo-0x10 + edx*2. EDX is the index register, with scale-factor 2. There is no base register. foo and -0x10 are both part of the displacement, both link-time constants. foo is a symbol address that the linker will fill in and subtract 0x10 from (because of the -0x10 assemble-time offset).

If you have the choice, use just a base instead of an index with a scale of 1. An index requires a SIB byte to encode, making the instruction longer. That's why compilers choose addressing modes like 8(%ebp) to access stack memory, not 8(,%ebp).

See also Referencing the contents of a memory location. (x86 addressing modes) for more about when you might use a base, and/or index, and/or displacement.

A 16-bit displacement is only encodeable in a 16-bit addressing mode, which uses a different format that can't include a scale factor, and has a very limited selection of which registers can be a base or index.

So a mode like 1234(%edx) would have to encode the 1234 as a 32-bit disp32 in 32-bit machine code.

Byte offsets from -128 .. +127 can use a short-form 8-bit encoding. Your assembler will take care of this for you, using the shortest valid encoding for the displacement.

All of this is identical in 64-bit mode for 64-bit addressing modes, with disp32 also being sign-extended to 64-bit just like disp8.