Home Can anyone explain to me what this shifting means under this circumstance: uint16_t register0 = (instruction >> 9) & 0x7

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Can anyone explain to me what this shifting means under this circumstance: uint16_t register0 = (instruction >> 9) & 0x7

byMR

December 3, 2022

I’m trying to write a VM (LC-3), and on this ADD instruction I encountered this statement. Basically the "register0" is the DR register, but I don’t really understand what is actually shifting and why 9. Also the AND operator with the 0x7 value.

|15|14|13|12|11|10|9|8|7|6|5|4|3|2|1|0|
|    0001   |  DR   | SR1 |0| 00| SR2 |

Could anyone please explain it to me in detail?

ADD {
    /* destination register (DR) */
    uint16_t r0 = (instr >> 9) & 0x7;
    /* first operand (SR1) */
    uint16_t r1 = (instr >> 6) & 0x7;
    /* whether we are in immediate mode */
    uint16_t imm_flag = (instr >> 5) & 0x1;
    if (imm_flag) {
        uint16_t imm5 = sign_extend(instr & 0x1F, 5);
        reg[r0] = reg[r1] + imm5;
    } else {
        uint16_t r2 = instr & 0x7;
        reg[r0] = reg[r1] + reg[r2];
    }
    update_flags(r0);
}

>Solution :

What it’s doing is isolating the 3 bits that represent the DR register so they become a standalone number.

Let’s say the entire sequence looks like this:

1110101101101011
    ^^^
     DR

Shifting 9 bits right gives this:

and & 0x7 (bitwise AND) isolates the 3 lowest bits:

Similar operations are performed to isolate the values of SR1 and the immediate mode flag. Depending on that flag, SR2 may also be required, but as it’s already in the lowest 3 bits, no shifting is needed.