The content of PC in the computer is 3A1 (ail numbers are in hexadecimal). The c
ID: 3765481 • Letter: T
Question
The content of PC in the computer is 3A1 (ail numbers are in hexadecimal). The content of AC is 7EC0. The content of memory address at address 3A1 is 932E. The content of memory at address 32E is 09AC. The content of memory at address 9AC is 7B9F. What is the instruction that will be fetched and executed next? Show the binary operation that will be performed in the AC when the instruction is executed? Give the contents of registers PC: AR: DR: AC: and IRin hexadecimal at the end of the instruction cycle.Explanation / Answer
It is given that,
The content of PC=3A1
The content of AC =7EC0
The content of memory address at address 3A1 is 932E.
The content of memory address at address 32E is 09AC.
The content of memory at address 9AC is 7B9F.
a)
Since the content of PC is 3A1, the next instruction to be executed is at memory location 3A1.
Thus the next instruction to be executed is 932E.
Therefore , Instruction register (IR)=932E and Address register (AR)= 32E.
b)
Observe the instruction 932E, in which 9 indicates that the instruction is indirect and the operation is binary addition .
The binary addition performed in the accumulated is as follows:
DR= M[M[AR]]= M[M[32E]]=M[9AC]= 7B9F [since, the instruction is indirect]
AC=AC+DR
=7EC0 + 7B9F
= FA5F
0111 1110 1100 0000
0111 1011 1001 1111
+
1111 1010 0101 1111= FA5F
c)
PC=3A2 [Since, after the instruction fetched, generally, PC is incremented by 1]
AR= 32E.
DR=7B9F
AC= FA5F [after the execution of binary addition, the result will be stored in AC]
IR=932E
0111 1110 1100 0000
0111 1011 1001 1111
+
1111 1010 0101 1111= FA5F
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