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An 8086 interrupt can take placedue to the following reasons:
1. Hardware interrupts caused by some external hardware device.
2. Software interrupts that can be invoked with the help of INT instruction.
3. Conditional interrupts that are mainly caused because of some error condition produced in 8086 by the execution of an instruction.
When an interrupt can be serviced by a procedure it is known as the Interrupt Service Routine (ISR). Starting addresses of the interrupt service routines are present in first 1K addresses of the memory. This table is known as the interrupt vector table.
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I think yes...so if 8259 gives interrupt then it will be serviced instantly but since into pin is left hanging the insr bit will never be set...and so whenever any interrupt occurs
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