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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.
ROL : Rotate Left without Carry: This instruction rotates the content of the destination operand to the left by the specified count bit-wise excluding the carry. The most signific
write an assembly language program to find average of odd numbers from an array of 8 bit numbers
Program Translation Sequence Developing a software program to accomplish a particular task, the implementer chooses an appropriate language, develops the algorithm (a sequence
add the contents of the defined memory locations 120, 133, 122 using mov instruction in dosbox
Example : Add the contents of the 2000H: 0500H memory location to contents of 3000H: 0600H and store the result in 5000H: 0700H. Solution : Unlike the past example progra
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Computes the integral square root: Problem: Square Root: For this problem you will write a short assembly program that computes the integral square root of an input numb
Write an assembly language program that will: accept keyboard input of a positive integer value N; compute the sum S= 1+ 2 + 3 + ... + N; print (output) the computed su
SBB: Subtract with Borrow :- The subtract with borrow instruction subtracts the source operand and the borrow flag (CF) which might reflect the result of the past calculations,
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