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HW11.3. Executing Store Instructions (3 Instructions) An LC-3 computer starts with the following register and memory contents: Registers RO: X30EB R4: x24BF R1: x2F6E R5: x2FF3 R2: x16B7 R6: x63D1 Memory x3000: ~3404 x3001: X7351 x3002: XB600 x3003: x3003 R3: X30BE R7: x2F7D PC: X3000 x3004: x3002 All other memory locations initially contain x0000. The LC-3 computer starts execution at PC = x3000 and continues until the instruction at x3002 finishes executing. You may find it useful to consult the LC-3 Handout. What values are at the following memory locations? Enter your answers in hexadecimal. x3003: X3004: x3005: X3006:

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The values for given memory locations are  x30BE, x2F6E, x16B7 , x0000.

What are he steps followed to get the memory locations?

ST

This is Store instruction which store the value in source register to address obtained by adding sign extended 9 bit offset with incremented PC.

Syntax-

ST Rs, Offset

Instruction format

0011  Rs 9 bit offset

STI

This is store indirect instruction which store the value in source register at address which is at address obtained by adding sign extended 9 bit offset with incremented PC.

Syntax-

STI  Rs, Offset

Instruction format

1011  Rs 9 bit offset

STR

This is store instruction which store the value in source register at address obtained by adding sign extended 6 bit offset with base register.

Syntax-

STR Rs, Rb,Offset

Instruction format

0111 Rs Rb 6 bit offset

Consider instructions at x3000 to x3002

x3404 = 0011 0100 0000 0100

Opcode = 0011, so instruction is ST.

Instruction format

0011 Rs 9 bit offset

Rs = 010 = R2

Offset = 0 0000 0100 = 4

Instruction will be,

ST R2,#4

Store R2 at address PC+1+4  = x3001+4 = x3005.

So, [x3005] = x16B7

x7351 = 0111 0011 0101 0001

Opcode is 0111, so instruction is STR

Instruction format

0111 Rs Rb 6 bit offset

Rs = 001 = R1

Rb = 101 = R5

Offset = 010001 = 17

So, instruction is ,

STR R1,R5,#17

Store, R1 at R5+17.

R5+17  = x2FF3 +x0011 = x3004  

So, R1 is stored at x3004.

[x3004] = x2F6E

xB600 =1011 0110 0000 0000

Opcode = 1011, so instruction is STI

Instruction format

1011 Rs 9 bit offset

Rd = 011 = R3

Offset = 000000000

So, instruction is,

STI R3,#0

Therefore, store R3 at address which is stored at address PC+1+0 = x3003. Value at x3003 is x3003. So, R3 stored at x3003.  

So, [x3003] = x30BE

So, values in locations,

x3003 = x30BE

x3004 = x2F6E

x3005 = x16B7  

x3006 = x0000

To know more about memory locations refer:

https://brainly.com/question/14447346

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