Consider the following sequence of micro operations :

MBR ← PC

MAR ← X

PC ← Y

MEMORY ← MBR

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

Correct Answer: Option 4) Initiation of interrupt service

Key Points

Let's examine the steps one by one:

1. MBR ← PC:
   • The content of the Program Counter (PC), which holds the address of the next instruction to be executed, is copied to the Memory Buffer Register (MBR).
   • This is typically done to save the return address before jumping to an interrupt service routine (ISR).
2. MAR ← X:
   • The Memory Address Register (MAR) is loaded with the address X.
   • This is the location where the return address (in MBR) will be stored.
3. PC ← Y:
   • The Program Counter is updated to Y, which points to the start of the ISR.
   • This redirect of the control flow to a new memory location is characteristic of interrupt handling.
4. MEMORY ← MBR:
   • The value in the MBR (the original PC value) is stored in memory at the address specified by MAR (which is X).
   • This effectively saves the return address so that the CPU can resume the original program after servicing the interrupt.

Therefore, this sequence of micro-operations matches the typical behavior during the initiation of an interrupt service routine, where:

• The current execution state (PC) is saved,
• The PC is redirected to the ISR location.

Final Answer: Option 4) Initiation of interrupt service