Hello I can’t figure out why in the add instruction I need to and by 7 this is the cpp code for the Add instruction

uint16_t dr = (instr >> 9) & 0b111;
                uint16_t sr1 = (instr >> 6) & 0b111;
                uint16_t sr2 = instr & 0b111;

                uint16_t second = registers[sr2];
                uint16_t immediateFlag = (instr >> 5) & 0b1;
                if (immediateFlag) {
                    uint16_t imm5 = instr & 0b11111;
                    second = signExtend(imm5, 5);
                }

                registers[dr] = registers[sr1]   second;

all the lines anded with 7 are the parts I don’t get. This is how the instruction looks like:

• bits 15-12 opcode(0001)
• bits 11-9 destination register
• bits 8-6 source1
• bits 5 0 or 1 (immediate mode)
• bits 4-3 nothing
• bits 2-0 source2

How does this 0b111 (7 in decimal) come into play and why?

CodePudding user response：

Take a look at the first line of code: it tries to decode the destination register, which is in bits 9-11 of your input number.

Assuming instr has 16 bits abcdefgh ijklmnop, then we want to extract bits 9-11, which is efg:
instr >> 9 shifts everything to the right by 9 bits, but the answer still has 16 bits: 00000000 0abcdefg.
& 0b111 is a shorthand for & 0b00000000 00000111, so applying that to instr >> 9 results in 00000000 00000efg, or exactly the three bits we were hoping to extract.