Determining how many clock cycles AVR assembly language code will take to execute

Question

If I was given Assembly code like the below, how could I determine how many clock cycles it will take to execute?

      ldi r20, 250
loop: inc r20
      brne loop
      nop

In the datasheet, all the instructions take up 16 bits (1 instruction word).

Attempting it myself, I get 14 as the answer. Since ldi r20, 250 is called once (1 cycle), then the loop is called 6 times before overflow to zero occurs (6x2=12 cycles). And finally at the end, nop takes 1 cycle. In total that is 14 cycles.

However, the answer is apparently 19 cycles. Would anyone be able to tell me what I have done wrong?

Answer 1

You've skipped including the inc at each loop. So each loop will be brne (2) + inc (1). Therefore your calculation should be (r20 value in brackets):

1
1 (251)
2
1 (252)
2
1 (253)
2
1 (254)
2
1 (255)
2
1 (0)
1
1

The brne is 1 cycle when the branch is not taken.

Its a bit more visible if you unroll that loop:

; (4 cycles)
ldi r20, 250 ; 1
inc r20      ; 1
nop          ; these two represent the brne at 2 cycles because it branches
nop

; (3 cycles)
inc r20      ; 1
nop          ; these two represent the brne at 2 cycles because it branches
nop

; (3 cycles)
inc r20      ; 1
nop          ; these two represent the brne at 2 cycles because it branches
nop

; (3 cycles)
inc r20      ; 1
nop          ; these two represent the brne at 2 cycles because it branches
nop

; (3 cycles)
inc r20      ; 1
nop          ; these two represent the brne at 2 cycles because it branches
nop

; (2 cycles)
inc r20      ; 1
nop          ; this represents the brne at 1 cycle, because its just overflowed and therefore ** will not branch **

; (1 cycle)
nop