Why does this code use multiple shifts instead of one?

Question

I came across the following piece of 16-bit x86 code for multiplying a value by 40, using just shifts and additions:

; BX holds the value we want to multiply.
; The result is stored in AX.
MOV AX, BX

; Multiply by 4 using two shifts
SHL AX, 1
SHL AX, 1

; Add the original value, this gives us BX * 5
ADD AX, BX

; Now multiply by 8 using three shifts for the final result
SHL AX, 1
SHL AX, 1
SHL AX, 1

Now what I'd like to know is why this code uses multiple shifts in a row instead of just doing SHL 2 and SHL 3. It was almost certainly written by hand, so I assume there was some speed benefit or something. Does anyone have any insights?

The code was written in 1991 and was targeting 286 and 386 class machines.

Answer 1

Shift/rotate with an immediate byte didn't exist until the 80186. On the 8086 only shifting/rotating by 1 or CL was possible. So it is likely you have some 8086 code.

Answer 2

There are two reasons I can think of and that has to do with overflow. Perhaps SHL AX, 1 and SHL AX 3 deal with overflow differently? otherwise, I don't see a difference unless there is possible some speed difference.