Assume the following shows the initial contents of the specified registers:REGISTER CONTENTSR1 0000 0000 0000 1001R2 0000 0000 0000 0101R3 0000 0000 0000 0010Also assume the following LC-3 machine instructions are loaded into memory at addresses shown:0x4000 1001 011 010 1 111110x4001 0001 011 011 1 000010x4002 0001 010 001 0 00 0110x4003 0000 001 001100000After the code above completes execution, what is the final value in each register below?(Enter your 16-bit answer in hex like the following example: 0x2A3F)A.) R1? B.) R2? C.) R3? D.) PC?
Question
Assume the following shows the initial contents of the specified registers:REGISTER CONTENTSR1 0000 0000 0000 1001R2 0000 0000 0000 0101R3 0000 0000 0000 0010Also assume the following LC-3 machine instructions are loaded into memory at addresses shown:0x4000 1001 011 010 1 111110x4001 0001 011 011 1 000010x4002 0001 010 001 0 00 0110x4003 0000 001 001100000After the code above completes execution, what is the final value in each register below?(Enter your 16-bit answer in hex like the following example: 0x2A3F)A.) R1? B.) R2? C.) R3? D.) PC?
Solution
The LC-3 machine instructions are as follows:
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0x4000: NOT R2, R1: This instruction performs bitwise NOT operation on the contents of R1 and stores the result in R2. The initial value of R1 is 0x0009. The bitwise NOT of 0x0009 is 0xFFF6. So, after this instruction, R2 = 0xFFF6.
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0x4001: ADD R3, R2, #1: This instruction adds the contents of R2 and 1, and stores the result in R3. The initial value of R2 is 0xFFF6. So, 0xFFF6 + 1 = 0xFFF7. So, after this instruction, R3 = 0xFFF7.
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0x4002: ADD R1, R3, R2: This instruction adds the contents of R3 and R2, and stores the result in R1. The values of R3 and R2 are 0xFFF7 and 0xFFF6 respectively. So, 0xFFF7 + 0xFFF6 = 0x1FFED. But since the register can only hold 16 bits, we only keep the lower 16 bits, so R1 = 0xFFED.
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0x4003: BRp R1: This instruction checks the condition codes set by the previous instruction and branches to the address specified by R1 if the previous result was positive. Since the previous result was not positive, the program counter (PC) is not changed and remains at the next instruction, which is 0x4004.
So, the final values in the registers are:
A.) R1 = 0xFFED B.) R2 = 0xFFF6 C.) R3 = 0xFFF7 D.) PC = 0x4004
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