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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.
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Instruction set of 8086 : The 8086/8088 instructions are categorized into the following major types. This section describes the function of each of the instructions with approp
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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 locati
DAA: Decimal Adjust Accumulator:- This instruction is utilized to convert the result of the addition operation of 2 packed BCD numbers to a valid BCD number. The conclusion has to
Memory Mapped I/O Memory I/O devices are mapped into the system memory map with ROM and RAM. To access a hardware device, simply write or read to those 'special' addresse
Control Transfer or Branching Instruction Control transfer instructions transfer the flow of execution of the program to a new address specified in the instruction indirectly o
I need to estimate the value of a definite integral using Riemann Sums and For our estimation let f(x) = x2 ,a=0, b=10 and n=5. Where a is the lower bound, b is the upper bound and
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