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Pointer and Index Registers
The pointers contain offset within the specific segments. The pointers BP, IP and SP generally containoffsets within thedata, code and stack segments respectively. The index registers are utilized as general purpose registers and for offset storage in case of indexed, and relative index andbased indexed addressing modes. The register SI is usually used to store the offset of source data in data segment while the register DI is utilized to store the offset of destination in data or extra segment. The index registers are specifically useful for string manipulations.
Flag Register : 8086 has a 16-bit flag register which is divided into 2 parts, viz. (a)machine control flagsand (b)condition code or status flags. The condition code flag regi
END : END of Program:- The END directive marks the ending of the assembly language program. When the assembler comes across this END directive, it avoided the source lines avai
PTR : Pointer:- The pointer operator which is used to declare the type of a variable, label or memory operand. The operator PTR is prefixed by either WORD or BYTE. If the prefi
#Write a function to calculate the following arithmetic operation and return the result. A = 2 + (3x)3 + y/2n (x, y and n are arguments of the function where x is an integer in the
* * * * **** * * * * * How can i print this help me pls
I need a division subroutine. Asks for two inputs, then displays the inputs and shows the answer with a remainder. Mine isnt displaying the inputs correctly.
Write the structure of For…Next loop in VB.Net and also write a program to print integers from 1 to 10 on the console.
assempbly language routine that takes an array named A containing n bytes of postive numebrs and fills two arranys, array B containing n words and array C containing n long words
RISC Characteristics : The concept of RISC architecture include an attempt to reduce execution time by make simple the instruction set of the computer. The main c
from pin description it seems that 8086 has 16 address/data lines i.e.AD0_AD15.The physical address is however is larger than 2^16.How this condition can be handled
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