Write Assembly language program To subtract 2 16-bit BCD numbers

; HL = 2435
; DE = 6578


LHLD 6578H ;
XCHG ; DE = 6578
LHLD 2435H ; HL = 2435
MOV A,L ; 
SUB E ;
DAA ;
STA 2000H ;
MOV A,H ;
SBB D ;
DAA ;
STA 2001H ;
HLT ;

Write Assembly language program To subtract 2 8-bit BCD numbers

; 2000H = 24
; 2001H = 43


LDA 2000H ;
MOV B,A ;
LDA 2001H ;
SUB B ;
DAA ;
STA 2002H ;
HLT ;

Write Assembly language program To add 2 16-bit BCD numbers

; HL = 2435
; DE = 6578
; METHOD 1


LHLD 6578H ;
XCHG ; DE = 6578
LHLD 2435H ; HL = 2435
MOV A,L ; 
ADD E ;
DAA ;
STA 2000H ;
MOV A,H ;
ADC D ;
DAA ;
STA 2001H ;
HLT ;

Write Assembly language program To add 2 8-bit BCD numbers

; 2000H = 78
; 2001H = 34
; 2002H = 78 + 34 = 112
; METHOD 1


LXI H, 2000H ;
MOV A,M ;78
INX H ;34
ADD M ;78 + 34
DAA ;112
INX H ;
MOV M,A ;STORE 112
HLT ;

Write Assembly language program for 2-Digit BCD to Binary conversion

; NEED TO USE CONCEPT OF POSITIONAL 
; WEIGHTING
; EXAMPLE: 34
; 34 = 3X0A + 4
; 34 = 1E + 4
; 34 = 22
; 2000H = 34H
;
;
LDA 2000H ;
ANI F0 ; A = 30
RRC ; MAKE MSB LSB
RRC
RRC
RRC
MVI B, 0AH ; B = 0A
CONTI ADD A ;
DCR B ;
JNZ CONTI ;
MOV B, A ; B = 1E = A
LDA 2000H ;
ANI 0F ; A = 4
ADD B ; A = 22
STA 2001H ; STORE 22
HLT

Write a Subroutine for 8085 to generate delay 0f 10ms(assume .333us clock cycle)

; REQUIRED T STATES
; = 10ms/.333us
; = 30030 T STATES
; SO WE CAN SAY VALUE OF COUNT HERE WILL BE HIGH
; THEREFORE TAKING A REGISTER PAIR AS A COUNTER


LXI B, COUNT ; 10T LOWER ADDRESS IN B HIGHER IN C
L1 DCX B ; 6T
MOV A,B ; 4T MOVE LOWER ADDRESS IN A
ORA C ; 4T SEE IF BOTH A AND C ARE ZERO
JNZ L1 ; 10/7T
RET ; 10T
; Td = 10 + COUNT X ( 6 + 4 + 4 + 10) + 10 - 3 = 30030
; 24COUNT = 30030 - 17 = 30013
; COUNT = 1250(DEC) = 04E2(HEX)

Write a Subroutine for 8085 to generate delay 0f 100us(assume 320ns clock cycle)

; clock cycle given is 320ns
; therefore required T-states = 100us/320ns
;      = 312.5 Tstates


MVI C, COUNT ; 7 T STATES
L1 DCR C ; 4T
JNZ L1 ; 10/7 T
RET ; 10 T REMEMBER A SUBROUTINE IS TO BE MADE


; Td = 7(MVI C, COUNT) + (COUNT X (4+10)) + 10(RET) - 3(LAST JUMP WILL NOT OCCUR) = 312.5
; THEREFORE 14COUNT = 297.5
; COUNT = 21.32 = 21(DEC) = 15(HEX)

Program to find mINImum number in an array

LXI H, 2000H ; INITIALIZE STARTING ADDRESS
MVI C, 10H ; SIZE OF ARRAY IS 10
XRA A ;
BACK CMP M ; COMPARE WITH MEMORY
JC SKIP ;
MOV A, M ; IF NUMBER< MAX INTERCHANGE
SKIP INX H ;
DCR C ;
JNZ BACK ;
STA 1FFFH ;
HLT ;

Calculate sum of series of even numbers

; MEANS ADD ONLY EVEN NUMBERS IN AN ARRAY
;
;
;


MVI C, 05H ; SIZE OF ARRAY
LXI H, 2000H ; START ARRAY
MVI B, 00H ; SUM =0
BACK MOV A, M ;
ANI 01H ; MASK BITS D1 TO D7 LEAVING D0 ALONE
JNZ SKIP ;
ADD M ; ADD EVEN NUMBER
MOV B,A ;
SKIP INX H ;
DCR C ;
JNZ BACK ;
STA 1FFFH ;
HLT ;

Count the number of 1's in register

; EXAMPLE H = 0101 0010 = 52 NUMBER OF 1'S =3
;
;


MVI C, 00H ; nUMBER OF 1'S
MVI B, 08H ; AS THERE ARE ONLY 8 BITS IN REGISTER
LXI H, 2000H ;
MOV A, M ;
L2 RAR ; ROTATE THROUGH CARRY
JNC L1 ; IF CARRY 1 CONTINUE
INR C ;
L1 DCR B ;
JNZ L2 ;
MOV A, C ;
STA 1FFFH ;
HLT ;