128-bit shifts using assembly language?

Question

What is the most efficient way to do 128 bit shift on a modern Intel CPU (core i7, sandy bridge).

A similar code is in my most inner loop:

u128 a[N];         


        

Answer 1

In this particular case you could use a combination of x86 SHR and RCR instructions:

; a0 - bits 0-31 of a[i]
; a1 - bits 32-63 of a[i]
; a2 - bits 64-95 of a[i]
; a3 - bits 96-127 of a[i]
mov eax, a0
mov ebx, a1
mov ecx, a2
mov ecx, a3

shr eax, 1
rcr ebx, 1
rcr ecx, 1
rcr edx, 1

; b0 - bits 0-31 of b[i] := a[i] >> 1
; b1 - bits 32-63 of b[i] := a[i] >> 1
; b2 - bits 64-95 of b[i] := a[i] >> 1
; b3 - bits 96-127 of b[i] := a[i] >> 1
mov b0, eax
mov b1, ebx
mov b2, ecx
mov b3, edx

shr eax, 1
rcr ebx, 1
rcr ecx, 1
rcr edx, 1

; c0 - bits 0-31 of c[i] := a[i] >> 2 = b[i] >> 1
; c1 - bits 32-63 of c[i] := a[i] >> 2 = b[i] >> 1
; c2 - bits 64-95 of c[i] := a[i] >> 2 = b[i] >> 1
; c3 - bits 96-127 of c[i] := a[i] >> 2 = b[i] >> 1
mov c0, eax
mov c1, ebx
mov c2, ecx
mov c3, edx

If your target is x86-64 this simplifies to:

; a0 - bits 0-63 of a[i]
; a1 - bits 64-127 of a[i]
mov rax, a0
mov rbx, a1

shr rax, 1
rcr rbx, 1

; b0 - bits 0-63 of b[i] := a[i] >> 1
; b1 - bits 64-127 of b[i] := a[i] >> 1
mov b0, rax
mov b1, rbx

shr rax, 1
rcr rbx, 1

; c0 - bits 0-63 of c[i] := a[i] >> 2 = b[i] >> 1
; c1 - bits 64-127 of c[i] := a[i] >> 2 = b[i] >> 1
mov c0, rax
mov c1, rbx

Update: corrected typos in 64-bit version