Question

Determine exactly how many instruction cycles the delay loop in the code below (between the comment lines Begin Delay Loop and End Delay Loop) takes as a function of the variables Count1 and Count2.

**Please Explain**

MaxCount EQU H'0A Equates are good for defining literals CBLOCK H'20 Count1 Count2 Scratch A CBLOck defines a sequential regio:n ; Count1 is in location H'20 ; Count2 is in location H'21 ; Scratch is in lication H'22 ; ENDC ends the definition block ENDC ORG 0x0000 GOTO Init Init ORG 0x0008 BSF STATUS, RPO CLRF TRISB CLRF ANSEL BCF STATUS, RPO SetScr MOVLW H'01' MOVWF Scratch MainLoop MOVF Scratch, W MOVWF PORTB Begin Delay Loop MOVLW Count2 MOVWF Index2 Loop MOVLW Count1 MOVWF Index1 SubLoop NOP NOP DECFSZ Index1, F GOTO SubLoop DECFSZ Index2, F GOTO Loop End Delay Loop RRF Scratch, F ANDWF Scratch, F BTFSC STATUS, Z GOTO SetScr GOTO MainLoop Finish END

Answer 1

Answer :- CBLOCK in PIC assembly has the same meaning as the ENUM in C language. So the value held by variable Count1 and Count2 are 32 and 33 respectively(decimal value).

The MOVLW, MOVF and NOP instructions take 1 MC,
GOTO instruction takes 2 MC,
DECFSZ takes 1 MC if result is not zero, else it takes 2 MC.

Begin Delay Loop MOVLW Count2 1MC MOVWF Index2- 1 MC Loop MOVLW MOVWF Index1 Count 1 2 MC for 33 times = 66 MC SubLoop NOP 2 NOPs runs for 32 x 33 times, giving 2 x 32 x 33 MC 2112 MC NOP DECFSZ Index1, F (31+2) MC for each 33 Count2 value 1089 MC GOTO SubLoop (31 x 2) MC for each 33 Count2 or Index2 value 2046 DECFSZ Index2, F (32+2 MC since Index2 has value 33, giving 34 MC GOTO Loop End Delay Loop This GOTO runs for 32 times only, so giving 32 x 2-64 MC Thus total machine cycle count is = (2 + 66 + 2112 + 1089 + 2046 + 34 + 64) MC-5413 MC