Test by Bikram | Computer Organization and Architecture | Test 2 | Question: 9

Question

A two word instruction is stored in memory at an address designated by symbol $S$. The address field of the instruction (stored at $S+1$) is designated by symbol K. The operand used during the execution of the instruction is stored at an address symbolized by the $P$. Which of the following is/are correct if the addressing mode used is the relative addressing mode?

• A. $P = M[ M[S] ] + K$
• B. $P = M[K+1] + S$
• C. $P = M [S ] + (K + 2)$
• D. $P = (S + 2) + M [ K ]$

Answer 1

 question is saying about the 2-word length and yes it has clearly mentioned that it is the relative mode of addressing. 

  relative addressing is just for displacement means it transfers the control to its effective address.

   

    example: jump, -24  ,

   +----+   ------------------------------+
    | jump |           offset                |          
   +----+   ------------------------------+
    

        S: jump
     S+1: -24  here it is given that K = s+1 , while excuting this instruction pc= k+1;

   (Effective PC address = next instruction address + offset, offset may be negative)
and it's given that effective address (address, where the operand is present ) is P.

so, p = pc +M(s+1)

         =( k +1) +M(k)

        =(s+1+1) + M(k)

      so, (S+2) + M[K]   option D

Comments

 The address field of the instruction (stored at S+1) is designated by symbol K

Isn't the above statement should instead be "The address of the address field of the instruction (stored at S+1) is designated by symbol k". 

Answer 2

@bikram sir

explain sir not getting answer.

Comments

@hem chandra joshi

In Relative address mode  Effective Address = PC value + Address part

so P = (K + 1) + M[K] ; where M[K] = memory address of K . As PC store the address of next instruction to be executed so PC value is (K+1)

Since K = S + 1 is given in the question;

Hence P = ((S + 1)+1) + M[K]

             = (S+2) + M[K]

which is option D .

Answer 3

Given:

Operand used during the execution of the instruction is stored at an address P.

Instruction is stored in memory at an address S .

The address field of the instruction (stored at S+1) is designated by symbol K means K = S+1 .

In Relative address mode we calculate Effective Address = PC value + Address part

so P = (K + 1) + M[K] ; where M[K] = memory address of K .

Since K = S + 1 is given in the question;

Hence P = ((S + 1)+1) + M[K]

             = (S+2) + M[K]

Comments

sir instruction is 2 word long

so let instruction is ADD R1,R2(X)

AND IF THIS INSTRUCTION IS STORE AT ADDRESS S  BUT IS TWO WORD LONG SO WILL IT NOT OCCUPY S AND S+1 MEMORY LOCATIONS?

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@hem chandra

Yes, when instruction is 2 word long , and instruction store at address S , it occupy S and S+1 memory location.

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Sir why di u do k+1 . K is already address of next instruction right ? then why k+1

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why (S+2) + M[K]? why not (S+2) + K ?