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Program : A program to move a string of the data words from offset 2000H to offset 3000H the length of the string is OFH.
Solution :
For writing this program, we will use an significant facility, available in the 8086 instruction set, for example move string byte/word instruction. We will also study the flexibility imparted by this instructions to the 8086 assembly language program. First write the Program for 8085, supposing that the string is available at location 2000H and is to be moved at 3000 H.
Now consider DS is suitably set, write the sequence for 8086. At first by using the index registers, the program may be written as given.
BY Comparing the above listings for 8085 and 8086 we can conclude that every instruction in 8085 listing is replaced by an alike instruction of 8086. The above 8086 listing is completely correct but it is not competent. Let us try to write the listings for the similar purpose by using the string instruction. Due to the assembler directives and the syntax, one may sense that the program is lengthy, though it eliminates 4 instructions for a MOVSW instruction.
General Data Registers Given figure indicate the register organization of 8086. The registers DX, CX, BX and AX are the general purpose 16-bit registers. AX is behaved as 16-bi
Port Mapped I/O or I/O Mapped I/O I/O devices are mapped into a separate address space. This is generally accomplished by having a different set of signal lines to denote a mem
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The 486 Introduced in the year 1989 the 80486 did not feature any radically new processor technology. Instead, it joints a 386 processor, a cache memory controller and a math c
Could I get an estimate on how much it would cost to write a program in assembly language?
Ask 2. Exchange higher byte of AX and higher byte of BX registers by using memory location 0160 in between the transfer. Then stores AX and BX registers onto memory location 0174 o
$NOMOD51 $NOSYMBOLS ;***************************************************************************** ; Spring 2013 Project ; ; FILE NAME : Project.ASM ; DATE : 3/30/20
Code for Reading Flow & Generating LED Output The code starts with the scanning of the PORT 3, for reading the flow status to check for various flow conditions and compare to
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
Example : Add the contents of the 2000H: 0500H memory location to contents of 3000H: 0600H and store the result in 5000H: 0700H. Solution : Unlike the past example progra
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