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INT N : Interrupt Type N:-
In the interrupt structure of 8086/8088, 256 interrupts are distinct equivalent to the types from OOH to FFH. When an instruction INT N is executed, the TYPE byte N is multiplied by value 4 and the contents of IP and CS register of the interrupt service routine will be taken from hexadecimal multiplication (Nx4) as offset address and 0000 as the segment address. In other terms, the multiplication of type N by value 4 (offset) points to a memory block in the 0000 segment, which have the IP and CS register values of the interrupt service routine.
For the execution of this instruction, the IF ought to be enabled.
Example :
Therefore the instruction INT 20H will find out the address of the interrupt service routine as follows:
INT 20H
Type* 4 = 20 * 4 = 80H
Pointer to CS and IP of the ISR is 0000: 0080 H
Given figure shows the arrangement of CS and IP register addresses of the ISR in the interrupt vector table.
NOT : Logical Invert: The NOT instruction complements (inverts) the contents of an a memory location or operand register bit by bit. The instance are as following: Example :
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hi!im looking for someone who expert in an assembly language and help me write the programmed!Thank you
#question.flow chart for a program to find out the number of even and odd numbers from a given series of 16-bit hexadecimal numbers.
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