问题
I learned about one address, two address, and three address instruction, but now I'd like to know, what kind of address instruction does x86 use?
回答1:
x86 is a register machine, where at most 1 operand for any instruction can be an explicit memory address instead of a register, using an addressing mode like [rdi + rax*4]. (There are instruction which can have 2 memory operands with one or both being implicit, though: What x86 instructions take two (or more) memory operands?)
Typical x86 integer instructions have 2 operands, both explicit, like add eax, edx which does eax+=edx.
Legacy x87 FP code uses 1-operand instructions, with the x87 stack, like faddp st1 where the top of the x87 stack (st0) is an implicit operand. SSE2 is baseline for x86-64, so it's no longer widely used.
Modern FP code uses SSE/SSE2 2-operand instructions like addsd xmm0,xmm1 or 3-operand AVX encodings like vaddsd xmm2, xmm0, xmm1
There are x86 instructions with 0, 1, 2, 3, and even 4 explicit operands.
There are multiple instruction formats, but explicit reg/memory operands are normally encoded in a ModR/M byte that follows the opcode byte(s). It has 3 fields:
- 2-bit Mode for the r/m operand (register direct
reg, register indirect[reg], [reg+disp8],[reg+disp32]). The modes with displacement bits signal that those bytes follow the ModR/M byte. - 3-bit r/m field (the register to use for that operand, or for memory addressing modes, an escape code that means there's a Scale/Index/Base byte after ModRM which can encode scaled-index addressing modes for the r/m operand). See rbp not allowed as SIB base? for the details of the special cases / escape codes.
- 3-bit reg field, always a register.
Most instructions are available in at least 2 encodings, reg/memory destination or reg/memory source. If the operands you want are both registers, you can use either opcode, either the add r/m32, r32 or add r32, r/m32.
Common instructions also have other opcodes for immediate source forms, but typically they use the reg field in ModR/M as extra opcode bits, so you still only get 2 operands like add eax, 123. An exception to this is the immediate form of imul added with 286, e.g. imul eax, [rdi + rbx*4], 12345. Instead of sharing coding space with other immediate instructions, it has a register dst and a r/m source in ModR/M plus the immediate operand implied by the opcode.
Some one-operand instructions use the same trick of using the reg field as extra opcode bits, but without an immediate. e.g. neg r/m32, not r/m32, inc r/m32, or the shl/shr/rotate encodings that shift by an implicit 1 (not by cl or an immediate). So unfortunately you can't copy-and-shift (until BMI2).
There are some special-case encodings to improve code density, like single-byte encodings for push rax/push rdx that pack the reg field into the low 3 bits of the opcode byte. And in 16/32-bit mode, one-byte encodings for inc/dec any register. But in 64-bit mode those 0x4? codes are used as REX prefixes to extend the reg and r/m fields to provide 16 architectural registers.
There are also instructions with some or all implicit operands, like movsb which copies a byte from [rsi] to [rdi], and can be used with a rep prefix to repeat that rcx times.
Or mul ecx does edx:eax = eax * ecx. One explicit source operand, one implicit source, and 2 implicit destination registers. div/idiv are similar.
Instructions with at least 1 explicit reg/mem operand use a ModR/M encoding for it, but instructions with zero explicit operands (like movsb or cdq) have no ModR/M byte. They just have the opcode. Some instructions have no operands at all, not even implicit, like mfence.
Immediate operands can't be signalled through ModR/M, only by the opcode itself, so push imm32 or push imm8 have their own opcodes. The implicit destinations (memory at [rsp], and RSP itself being updated to rsp-=8).
LEA is a workaround that gives x86 3-operand shift-and-add, like lea eax, [rdi + rdi*2 + 123] to do eax = rdi*3 + 123 in one instruction. See Using LEA on values that aren't addresses / pointers? The destination register is encoded in ModR/M's reg field, and the two source registers are encoded in the addressing mode. (Involving a SIB byte, the presence of which is signalled by the ModR/M byte using the encoding that would otherwise mean base = RSP).
VEX prefixes (introduced with AVX) provide 3-operand instructions like bzhi eax, [rsi], edx or vaddps ymm0, ymm1, [rsi]. (For many instructions, the 2nd source is the one that's optionally memory, but for some it's the first source.)
The 3rd operand is encoded in the 2 or 3-byte VEX prefix.
There are a few 3-operand non-VEX instructions, such as SSE4.1 variable blends like vpblendvb xmm1, xmm2/m128, <XMM0> where XMM0 is an implicit operand using that register.
The AVX version makes it non-destructive (with a separate destination encoded in the VEX prefix), and makes the blend-control operand explicit (encoded in the high 4 bits of a 1-byte immediate). This gives us an instruction with 4 explicit operands, VPBLENDVB xmm1, xmm2, xmm3/m128, xmm4.
x86 is pretty wild and has been extended many times, but typical integer code uses mostly 2-operand instructions, with a good amount of LEA thrown in to save instructions.
来源：https://stackoverflow.com/questions/53325275/what-kind-of-address-instruction-does-the-x86-cpu-have