wip: CPU implementation
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project-5-computer-architecture/hdl/CPU.hdl
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project-5-computer-architecture/hdl/CPU.hdl
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// This file is part of www.nand2tetris.org
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// and the book "The Elements of Computing Systems"
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// by Nisan and Schocken, MIT Press.
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// File name: projects/5/CPU.hdl
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/**
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* The Hack Central Processing unit (CPU).
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* Parses the binary code in the instruction input and executes it according to the
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* Hack machine language specification. In the case of a C-instruction, computes the
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* function specified by the instruction. If the instruction specifies to read a memory
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* value, the inM input is expected to contain this value. If the instruction specifies
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* to write a value to the memory, sets the outM output to this value, sets the addressM
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* output to the target address, and asserts the writeM output (when writeM = 0, any
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* value may appear in outM).
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* If the reset input is 0, computes the address of the next instruction and sets the
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* pc output to that value. If the reset input is 1, sets pc to 0.
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* Note: The outM and writeM outputs are combinational: they are affected by the
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* instruction's execution during the current cycle. The addressM and pc outputs are
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* clocked: although they are affected by the instruction's execution, they commit to
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* their new values only in the next cycle.
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*/
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CHIP CPU {
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IN inM[16], // M value input (M = contents of RAM[A])
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instruction[16], // Instruction for execution
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reset; // Signals whether to re-start the current
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// program (reset==1) or continue executing
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// the current program (reset==0).
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OUT outM[16], // M value output
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writeM, // Write to M?
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addressM[15], // Address in data memory (of M)
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pc[15]; // address of next instruction
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PARTS:
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// Instructions.
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// Decode the instruction bit.
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// i xx a cccccc ddd jjj
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// For C-Instruction the byte is stored as i(opcode) xx(notUsed) a cccccc(a/c = comp) ddd(destination) jjj(jump)
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// i xxxxxxxxxxxxxxx
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// For A-Instruction the i(opcode) is 0 and the rest is the 16(15)-bit value that is stored there.
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// Decode Instruction
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Not(in=instruction[15], out=isAInstruction); // Flip it and if 1 it's an A Instruction
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And(a=instruction[15], b=true, out=isCInstruction); // Check it with an And static true to check if both 1 then it's C instruction.
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// Don't need decoding for the rest when A instruction.
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// Just use instruction because instruction[15] will always be 0
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// Decode C & Create Wires
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// 1xA
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And(a=instruction[12], b=true , out=a1);
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// 6xC
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And(a=instruction[11], b=true , out=c1);
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And(a=instruction[10], b=true , out=c2);
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And(a=instruction[9], b=true , out=c3);
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And(a=instruction[8], b=true , out=c4);
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And(a=instruction[7], b=true , out=c5);
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And(a=instruction[6], b=true , out=c6);
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// 3xD
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And(a=instruction[5], b=true , out=d1);
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And(a=instruction[4], b=true , out=d2);
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And(a=instruction[3], b=true , out=d3);
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// 3xJ
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And(a=instruction[2], b=true , out=j1);
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And(a=instruction[1], b=true , out=j2);
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And(a=instruction[0], b=true , out=j3);
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// aluOut is not made yet
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// A Register
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// Should Load from ALU if
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And(a=isCInstruction, b=d1, out=loadAFromALU);
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Mux16(a=instruction, b=aluOut, sel=loadAFromALU,out=aRegIn);
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Or(a=isAInstruction, b=loadAFromALU, out=loadA);
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Register(in=aRegIn, load=loadA, out=aRegOut);
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// D Register
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And(a=isCInstruction, b=d2, out=loadD);
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Register(in=aluOut, load=loadD, out=dRegOut);
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// a - instruction12 - a1
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// a=0 → use A register
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// a=1 → use M (inM from memory)
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// ALU
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Mux16(a=aRegOut, b=inM , sel=a1 , out=aluYin);
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ALU(x=dRegOut,
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y=aluYin,
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zx=c1, nx=c2, zy=c3, ny=c4 , f=c5 , no=c6 ,
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out=aluOut ,
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zr=aluZrOut ,
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ng=aluNgOut
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);
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}
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