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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.
Opcode : The opcode generally appear in the first byte.but in a few instructions, a register objective is in the first byte and few other instructions may have their 3-bits of
Data copy/transfer Instructions MOV: This data transfer instruction transfers data from one register or memory location to another register or memory location. The source can
ADC: Add with Carry:- This instruction performs the similar operation a like ADD instruction, but adds the carry flag bit (which might be set as a result of the previous calculatio
Tabular comparison for µ PS' Parameters Tables (a) and (b) list the characteristic of Intel microprocessor. Table(a): Table(b): It has a 64 bit da
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
can u please give me ideas on Assembly Language Projects using Nasm
What is the hex for + and - under with a sum involved
Program : Write an assembly program to find out the number of positive numbers and negative numbers from a given series of signed numbers. Solution : Take the i th num
Program : Write a program to perform a one byte BCD addition. Solution : It is consider that the operands are in BCD form, but the CPU considers it as hexadecimal and acco
programs
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