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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.
what is double hashing
HELLO I AM TRYING TO ADD AND SUBTRACT BUT I SEEM CAN''T FIND THE CORRECT REGISTER TO PUT IN
Read Architecture: Look Through Main memory that located is conflicting the system interface. The least concerning feature of this cache unit is that it remain between the proc
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
Memory Address Decoding Binary Decoders - Decoders have 2n-inputs and n outputs, each input combination results in a single output line contain a 1, and all other lines contain
Trying to convert small programs from C to 8086 assembly language using emu 8086 emulator. I converted to low level C, but struggling with converting to the Assembly language.
1- Write an assembly program that: a- Defines an array of 10 (word type)elements; b- Finds out the number of negative elements c- Calculate the summation of the posi
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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.
Display control 8279 provides a 16 byte display memory and refresh logic. Every address in the display memory corresponds to a display unit with address zero represen
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