Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
MOVSW/MOVSB : Move String Word or String Byte: Imagine a string of bytes, stored in a set of consecutive memory locations is to be moved to another set of the destination locations. Starting byte of the source string is located in the memory location whose address can be computed by using DS (data segment) and SI (source index) contents. Starting address of the destination locations where this string has to be relocated is given by ES (extra segment) and Dl (destination index) contents. Starting address of the source string is 10H*DS+[SI], whereas the starting address of the destination string is 10H*ES+[DI]. The MOVSB/MOVSW instruction therefore, moves a string of bytes/ words pointed to by DS: SI pair (source) to the memory location pointed to by ES: Dl pair (destination). The REP instruction prefix is utilized with MOVS instruction to repeat it by a value given in the register counter (CX). The length of word string or byte string ought to be stored in register CX register. Flags are remaining unaffected by this instruction.
After the MOVS instruction is executed, the index registers are automatically updated and register CX is decremented. The decrementing or incrementing of the pointers, for example DI and Sl depend on the direction flag DF. If flag DF is 0, the index registers are incremented, or else, they are decremented, in all casa of the string manipulation instructions. Following string of instructions explain the execution of the MOVS instruction.
Example :
Write an application that does the following:(1) fill an array with 50 random integers; (2) loop through the array, displaying each value, and count the number of negative values;
$NOMOD51 $NOSYMBOLS ;***************************************************************************** ; Spring 2013 Project ; ; FILE NAME : Project.ASM ; DATE : 3/30/20
calculate the number of one bits in bx and complement an equal number of least significant bits in ax hint use the xor instruction
INTO : Interrupt on Overflow:- It is executed, when the overflow flag OF is set. The new contents of IP and CS register are taken from the address 0000:0000 as described in INT
Project Description: Write an 80x86 assembly program that performs the following functions: Reads a set of integers from a file into an array. The data file name is to be
I/O interface I/O devices such as displays and keyboards establish communication of computer with outside world. Devices may be interfaced in 2 ways Memory mapped I/O and I/
Execution Unit (EU) and Bus Interface Unit (BIU) : 8086 consist of two processors called EU and BIU. Two Processors can work parallel. This improves speed of execution. BIU fi
Segment Registers The 8086 addresses a segmented memory unlike 8085. The complete 1 megabyte memory, which 8086 is capable to address is divided into 16 logical segments.Thusea
Write a 32-bit program which when run, allows the user to select from a menu: (1) Enter a Binary Number (2) Enter a Decimal Number (3) Enter a Hexadecimal Number
Multiply two numbers by using shift and rotate instruction
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd