GATE CS 2013 Q.28 (CS Q.28): I/O Interface, Interrupts and DMA

Consider the following sequence of micro-operations.

\text{MBR} \leftarrow \text{PC}

\text{MAR} \leftarrow X

\text{PC} \leftarrow Y

\text{Memory} \leftarrow \text{MBR}

Which one of the following is a possible operation performed by this sequence?

Instruction fetch

Operand fetch

Conditional branch

Initiation of interrupt service

Official answer and solution

Official answer: D
From: Official GATE 2013 answer key

Solution

Identify the purpose of saving the program counter (PC) to memory and updating PC with a new target address Y.

Let us trace the given sequence of micro-operations step by step:

MBR <- PC: The current value of the program counter (holding the return address of the next instruction) is placed into the memory buffer register (MBR).
MAR <- X: The address $X$ (which specifies where the return address should be saved, such as the stack top or a designated interrupt return-address memory word) is loaded into the memory address register (MAR).
PC <- Y: The program counter is updated with address $Y$ , which is the starting address of the Interrupt Service Routine (ISR) or interrupt handler.
Memory <- MBR: The saved PC value in MBR is written into main memory at the location pointed to by MAR.

This exact sequence preserves the return address and branches control to an interrupt vector address $Y$ . Therefore, this sequence performs the initiation of interrupt service.

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2013 Computer Science and Information Technology (CS)
Question: 28 of 65, printed as CS Q.28
Section: Computer Science and Information Technology

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