Department of Electrical Engineering and Computer Science 6. (55 Points) For eac

Question

Department of Electrical Engineering and Computer Science 6. (55 Points) For each instruction below assume the initial value of the registers are as below: R0 = 0x00000100 RI = OxAABBCCDD R2 = 0x00000002 Oxo0000 10o 1011 102 33 103 4 104 55 LDR R1,[R0, R2, LSL#1] a. After the execution of the above instruction value of following registers: 106 77 07 88 108 9 R2 = b. RI,[R01.3 LDR RO = 0x00000C00 RI = 0xAABBCCDD R2 = 0x00000002 After the execution of the above instruction Ox 00000e A eol eD C02 ERF Co3 IA value of following registers: eob ID RI = Co8 IF R2 =

Explanation / Answer

I suppose that these questions belong to microprocessor. Pardon me if i am wrong

a)

LDR R1,[R0,R2, LSL #1]

LSL #1 - is logical shift left operation

thus u need to left shift the value of R2( the adjacent register or previous register) by 1

so the above operation can be inferred as

R1 = value at the address [R0+(R2*2)] where R0 and R2 value do not change. R1 value changes.

R1=*[0x00000100+(0x00000002 *2)]

R1=*[0x00000100+(0x00000004)]

R1=*[0x00000104]

thus R1= 55, R0=0x00000100, R2=0x00000002

b)

LDR R1,[R0], #3

in this operation first the effective address of the register R0 is loaded to R1 and then the value of R0 is cahnged to R0+3

thus R1=R0 then R0= R0 +3

R1=*R0

R1=*[0x00000C00]

R1=AB and R2=0x00000002, R0=0x00000C03

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