Question

The hypothetical machine of Figure 3.4 also has two I/O instructions:0011 = Load AC from I/O0111...

The hypothetical machine of Figure 3.4 also has two I/O instructions:

0011 = Load AC from I/O

0111 = Store AC to I/O

In these cases, the 12-bit address identifies a particular I/O device. Show the program execution (using the format of Figure 3.5) for the following program:

1. Load AC from device 5.

2.  Add contents of memory location 940.

3.  Store AC to device 6.

Assume that the next value retrieved fromdevice 5 is 3 and that location 940 contains a value of 2.

Example of Program Execution (contents of memory and registers in hexadecimal)

Program execution as per figure 3.5 from textbook:

Step 1: Instruction Register (IR) 3005.

Step 2: Accumulator (AC) 3

Step 3: Instruction Register (IR) 5940

Step 4: Accumulator (AC) 5 (3 +2)

Step 5: Instruction Register (IR) 7006

Step 6: Device 6 Accumulator (AC)

Explanation:

• Instruction register (IR) is the one which stores instruction currently being executed or decoded.

• Program counter (PC) is the one that holds address of the instruction being executed at current time.

• 12 bits is used to specify the address and 4 bits in the IR indicates accumulator to be loaded.

Let the contents of memory location be,

• 300 : 3005

• 301 : 5940

• 302 : 7006

Step 1:

In step1 the “Program Counter” holds the value of “300” and “Instruction register” holds contents of address “300” that is “3005”.

Step 2 & Step 3:

The content of address location “005” is moved to the accumulator and Instruction register holds the value “5940

Step 4:

Add contents of Accumulator with contents of address “940” and stored back to accumulator.

Step 5:

The next instruction (7006) is fetched from location 302 and PC is incremented.

Step 6:

The contents of the AC are stored in location 006

Thus, execution steps for the given problem according to figure 3.5 from text book is done.

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