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Assume the following shows the initial contents of the specified registers and memory locations:REGISTER CONTENTSR4 0101 0000 0110 0000R5 0011 0001 0000 0000R6 0010 1100 0001 0000ADDRESS CONTENTS0x5050 0011 0001 0010 00000x5051 0101 0000 0110 10110x5052 0111 0011 0000 1111Also assume the following LC-3 machine instructions are loaded into memory at addresses shown:0x5010 0110 101 110 0000100x5011 1010 100 001000000What is the effective address for each of the instructions above?(Enter your 16-bit answer in hex like the following example: 0x2A3F)A.)   first instruction above B.)   second instruction above

Question

Assume the following shows the initial contents of the specified registers and memory locations:REGISTER CONTENTSR4 0101 0000 0110 0000R5 0011 0001 0000 0000R6 0010 1100 0001 0000ADDRESS CONTENTS0x5050 0011 0001 0010 00000x5051 0101 0000 0110 10110x5052 0111 0011 0000 1111Also assume the following LC-3 machine instructions are loaded into memory at addresses shown:0x5010 0110 101 110 0000100x5011 1010 100 001000000What is the effective address for each of the instructions above?(Enter your 16-bit answer in hex like the following example: 0x2A3F)A.)   first instruction above B.)   second instruction above

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Solution

A.) The first instruction is a Load Base + Offset (LDR) instruction. The format is 0110 101 110 0000100. The base register (BR) is 110 (or R6 in decimal), and the offset6 is 0000100 (or 4 in decimal). The contents of R6 is 0010 1100 0001 0000 (or 0x2C10 in hexadecimal). Therefore, the effective address is the sum of the contents of the base register and the offset, which is 0x2C10 + 0x4 = 0x2C14.

B.) The second instruction is a Load Indirect (LDI) instruction. The format is 1010 100 001000000. The offset9 is 001000000 (or 0x100 in hexadecimal). The PC value after this instruction is fetched is the address of this instruction plus 1, which is 0x5011 + 1 = 0x5012. Therefore, the effective address is the sum of the PC value and the offset, which is 0x5012 + 0x100 = 0x5112.

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