Fetch: sending the address from the PC to the MAR
Name:
Date:
Class:
A
Quick Questions
Shade in the A or B circle for each question.
1. What happens at the very start of the fetch stage?
2. Which register message shows the first step of fetch?
3. After MAR <- PC, what does the PC hold?
4. Which bus carries the address from the MAR to RAM?
5. What does the PC hold at the start of fetch?
B
Match the Pairs
Draw a line connecting each item on the left to its match on the right.
| Left column | Right column | |
|---|---|---|
| 1MAR <- PC | AAt the start of fetch, the address in the PC is copied into the MAR | |
| 2MAR -> address bus -> RAM | BMAR <- PC | |
| 3The first step of fetch copies the address of the next instruction from the PC into the MAR | CThe PC is not emptied because the address is copied, not moved. The PC keeps its value at this point. | |
| 4After the copy, the PC still holds the same address because the value is copied, not moved | DThe address in the MAR travels along the address bus to the location in RAM | |
| 5MDR <- PC | EThe address from the PC must go to the MAR, not the MDR. The MDR holds data or instructions that come from RAM. | |
| 6MAR <- PC, then PC is emptied (PC = 0) | FPC = 100, MAR = 100 |
C
Definitions
MAR <- PC
D
Fill in the Blanks
Complete each line of code by filling in the missing parts.
1.
MAR ← __
2.
___ ← PC
3.
MAR _ PC
4.
MAR ← PC
// The address travels on the _______ bus
5.
PC = 250
MAR ← PC
// PC still holds ___
E
Spot the Error
Shade in the letter of the line that contains an error.
Question 1
A
PC <- 100
B
MAR <- PC
C
ADDRESS BUS <- MAR
D
PC is now empty
Question 2
A
PC <- MAR
B
Address bus sends MAR to RAM
C
PC keeps its address
Question 3
A
Fetch begins: MDR <- PC
B
MAR sends address on addr bus
C
PC still holds the address
Question 4
A
PC holds address 250
B
MAR <- PC
C
MAR holds 251
Question 5
A
PC holds 40
B
MAR <- PC + 1
C
MAR holds 40
D
Address bus carries 40
Answer Key
Fetch: sending the address from the PC to the MAR
Section A: Quick Questions
Q1
B — The address in the PC is copied into the MAR
Q2
B — MAR <- PC
Q3
A — It still holds the address, because it was copied
Q4
A — Address bus
Q5
A — The address of the next instruction to be fetched
Section B: Match the Pairs
Q1
MAR <- PC → At the start of fetch, the address in the PC is copied into the MAR
Q2
MAR -> address bus -> RAM → The address in the MAR travels along the address bus to the location in RAM
Q3
The first step of fetch copies the address of the next instruction from the PC into the MAR → MAR <- PC
Q4
After the copy, the PC still holds the same address because the value is copied, not moved → PC = 100, MAR = 100
Q5
MDR <- PC → The address from the PC must go to the MAR, not the MDR. The MDR holds data or instructions that come from RAM.
Q6
MAR <- PC, then PC is emptied (PC = 0) → The PC is not emptied because the address is copied, not moved. The PC keeps its value at this point.
Section C: Definitions
Q1
MAR <- PC — At the start of fetch, the address held in the PC is copied into the MAR. It then travels along the address bus to RAM. The PC still holds the address at this point, because it is copied, not moved.Section D: Fill in the Blanks
Q1
MAR ← PCQ2
MAR ← PCQ3
MAR ← PCQ4
MAR ← PC
// The address travels on the address busQ5
PC = 250
MAR ← PC
// PC still holds 250Section E: Spot the Error
Q1
D — The address is copied from the PC to the MAR, not moved. The PC still holds 100 at this point. The line should be: PC still holds 100.
Q2
A — The transfer goes the wrong way. At the start of fetch the address is copied from the PC into the MAR, so it should be: MAR <- PC.
Q3
A — The PC address goes into the MAR, not the MDR. The MDR holds data or instructions coming from RAM. It should be: MAR <- PC.
Q4
C — The address is copied unchanged, so the MAR holds 250, the same as the PC. The PC is incremented later in the fetch stage, not during the copy.
Q5
B — The MAR receives an exact copy of the PC. Adding 1 is the separate increment step that comes after. It should be: MAR <- PC.