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DW : Define Word:-
The DW directive serves the same purposes as the DB directive, but now it makes the assembler which reserves thenumber ofmemory words (16-bit) instead of bytes. Some instance aregiven to explain this directive.
Example :
WORDS DW 1234H, 4567H, 78ABH, 045CH,
This makes the assembler reserve 4 words in memory (8 bytes), and by initializing the words with the specified values in the given statements. At the time of initialisation, the lower bytes are stored at the lower memory addresses, whereas the upper bytes are stored at the higher addresses. Another option of the DW directive can be explained with the DUP operator.
WDATA DW 5 DUP (6666H)
This statement reserves five words, for example10-bytes of memory for a word label WDATA and initialises all the word locations having 6666H.
Opcode : The opcode generally appear in the first byte.but in a few instructions, a register objective is in the first byte and few other instructions may have their 3-bits of
Problem (a) Prepare the assembly code sequence for each of the four styles (accumulator, memory-memory, stack, load/store) of machine for the code fragment: A = B + C;
init_lcd ;(this initialises a 2 row lcd) bcf TRISA,0 ;PORTA bit 0 as an output (lcd RS pin) bcf TRISA,1 ;PORTA bit 1
add the contents of the defined memory locations 120, 133, 122 using mov instruction in dosbox
write a program using assembly language that print your name
#include"lcd.asm" ; assembly file is included for displaying lcd characters Main: PORTA EQU 0xF80 ; PORTS PORTB EQU 0xF81 PORTC EQU 0xF82 PORTD EQU 0xF83 R
HELLO I AM TRYING TO ADD AND SUBTRACT BUT I SEEM CAN''T FIND THE CORRECT REGISTER TO PUT IN
I need to generate a random number bby using 8086 assembly language
TEST : Logical Compare Instruction: The TEST instruction performs bit by bit logical AND operation on the 2 operands. Each bit of the result is then set to value I, if the equival
I have two homework assignments due in 10 hours for the x86 processor assembly language
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