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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.
8255 Programmable Peripheral Interface Intel's 8255 A programmable peripheral interface provides a nice instance of a parallel interface. As shown the interface have a control
Pin Description of 8086 The microprocessor 8086 is a 16-bit CPU available in 3 clock rates, for example 5, 8 and 10 MHz, packaged in a40 pin CERDIP or plastic package. The 8
assembly language program to find larges number in an array
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I have two homework assignments due in 10 hours for the x86 processor assembly language
Software Interrupts Software interrupts are the result of an INT instruction in an executed program. It may be assumed as a programmer triggered event that immediately stops e
#I submitted my assignment this morning and it is still processing. How long does it take?
DMA DMA stands for Direct Memory Access It is uses same Address/Data lines on ISA bus It controls the ISA bus instead of the processor ("bus master") Floppy
Architecture Of 8088 The register set of 8088 is accurately the same as in to 8086. The architecture of 8088 is also same to 8086 except for 2 changes; a) 8088 has 4-byte instr
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
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