This is the fetch instruction that the CPU takes for executing.
At reset, the 8085 starts at location 0000H. The INTR interrupt goes wherever the interrupt opcode says to go. The TRAP interrupt goes to 0024H. The RST5.5, RST6.5, and RST7.5 interrupts go to 002CH, 0034H, and 003CH respectively. The RST X interrupts goto 0000H plus 10H times X, i.e. 0000H, 0008H, 0010H, ..., 0038H.
(b) -> ffffh (c) -> fffeh fffeh -> sp
program:assume cs:code,ds:datacode segmentmov ax,datamov ds,axmov cl,countmov si,offset str1mov di,0003hback:mov al,[si]xchg [di],almov [si],alinc sidec didec cljnz backhltcode endsdata segmentstr1 db 01h,02h,03h,04hcount equ 02hdata endsendresult:input: str1 (ds:0000h) = 01h,02h,03h,04houtput: str1 (ds:0000h) = 04h,03h,02h,01h
This is my program, and it works with all no.s except multiples of 2. org 100h MOV CX,0000H MOV DS,CX MOV SS,CX MOV SI,5000H MOV DI,5002H MOV [ DS:SI ],10H MOV [ DS:DI ],20H MOV SP,600FH MOV BX,[ DS:SI ] CMP BX,[ DS:DI ] JZ E1 JC SMALL THIK: MOV BX,0001H OK: MOV AX,[ DS:SI ] MOV DX,0000H DIV BX CMP DX,0000H JZ L1 L2: INC BX CMP [ DS:DI ],BX JC HCF JMP OK SMALL: MOV AX,[ DS:DI ] MOV [ DS:DI ],BX MOV [ DS:SI ],AX JMP THIK L1: MOV AX,[ DS:DI ] DIV BX CMP DX,0000H JNZ L2 PUSH BX INC CX JMP L2 HCF: MOV AX,0001H AGAIN: POP BX MUL BX DEC CX JNZ AGAIN LCM: MOV BX,AX MOV AX,[ DS:SI ] MUL [ DS:DI ] DIV BX E1 : INC DI INC DI MOV [ DS:DI ],AX ret
I am assuming that you mean "how do you change the time from a 24 hour clock to a twelve hour clock." Just subtract 1200 hours from any time over 1200H. 0000H is midnight. 0100H is 1:00 am etc. up to noon. 1200H is Noon. 1300H is 1:00 pm (ie 1300 - 1200 = 100 = 1:00 pm). 1600H would be 4:00 pm (1600 - 1200 = 400 = 4:00 pm). 2200H is 10:00 pm (2200 - 1200 = 1000 = 10:00 pm).
The 8086 interrupt table is from 0000h to 03FFh for interrupt 0 through interrupt 255. It is common practice to design systems that use only the lower-numbered interrupts and then use the upper part of the interrupt table for code or data. For more information see: http://datasheets.chipdb.org/Intel/x86/808x/datashts/8086/231455-005.pdf
Lxi b, 0000h lhld 8000h xchg lhld 8002h dcx d l006: lda 8002h add l mov l, a lda 8003h adc h mov h, a jnc l013 l013: inx b dcx d mov a, d ora e jnz l006 shld 8006h mov l, c mov h, b shld 8004h hlt
Code for An Assembly Language Program to find 2's Complement of given binary number in Assembly LanguageData Segment num db 00000010B Data Ends Code Segment Assume cs:code, ds:data Begin: mov ax, data mov ds, ax mov es, ax mov ah, 0000h mov al, num NOT al mov bl, al adc al, 00000001B mov bl, al Exit: mov ax, 4c00h int 21h Code Ends End Begin
he was a mean person who lived with mean people in a mean castle on a mean hill in a mean country in a mean continent in a mean world in a mean solar system in a mean galaxy in a mean universe in a mean dimension
you mean what you mean
Mean is the average.
Mean