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IRET : Return from ISR:-
When an interrupt service routine is called, before transferring control to it, the IP, CS register and flag registers are stored in the stack to mention the location from where the execution is to be continued, after the ISR is executed. Hence, in the ending of each ISR, when IRET is executed, the values of IP, CS register and flags are retrieved from the stack to continue the execution of the main program. The stack is modified consequently.
LOOP : Loop Unconditionally:-
This instruction executes the part of the program from the address or label mention in the instruction up to the loop instruction, CX number of times. Following sequence describe the execution. On every iteration, register CX is decremented automatically. In other terms, this instruction implements JUMF IF NOT ZERO and DECREMENT COUNTER structure.
The execution proceeds in the sequence, after the loop is executed, CX number of times. Lf CX is already OOH, the execution continues in sequence. Flags are remaining unaffected by this instruction.
AAS: ASCII Adjust AL After Subtraction AAS instruction correct the result in the AL register after subtracting operation of two unpacked ASCII operands. The result is in unpacked
code to add two matrices
00h-1h
LDS/LES Instruction execution : LAHF : Load AH from Lower Byte of Flag: - This instruction loads the AH register with the lower byte of the flag register. This instruction ca
Description: LC3 allows input from keyboard and output to display on the screen. This lab will exercise the input/output capability using LC-3 Assembly language. Procedure
Interrupt When the CPU detects an interrupt signal, it stops activity of current and jumps to a special routine, known an interrupt handler. This handler then detects why the i
Assembly Language: Inside the 8085, instructions are really stored like binary numbers, not a very good manner to look at them and very difficult to decipher. An assembler is
Memory Segmentation : The memory in an 8086/8088 based system is organized as segmented memory. In this scheme, the whole physically available memory can be divided into a n
Write an assembly language program to find the maximum of: y = x 6 - 14x 2 + 56x for the range -2 ≤ x ≤ 4, by stepping one by one through the range. The program should in
Difference between div and idiv
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