How do I convert a number to two's complement in verilog?

Question

I am trying to design a 4-bit adder subtracter in verilog. This is only the second thing I have ever written in verilog, and I don't know all the correct syntax yet. This is the module I have so far:

module Question3(carryin, X, Y, Z, S, carryout, overflow);
    parameter n = 4;
    input carryin, Z;
    input [n-1:0]X, Y;
    output reg [n-1:0]S;
    output reg carryout, overflow;

    if(Z==0)
    begin
     Y = not(y) + 4'b0001;
    end

    always @(X, Y, carryin)
     begin
      {carryout, S} = X + Y + carryin;
      overflow = carryout ^ X[n-1]^Y[n-1]^S[n-1];
     end

endmodule

My compiler (xilinx 10.1), keeps saying "Syntax error near if." I have tried many different ways of doing the conversion, including just using a Case that takes Y as an argument, then checks all the possible 4-bit combinations, and converts them to two's complement.

Z is what determines if the adder does subtraction or addition. If it's 0, it means subtraction, and I want to convert y to two's complement, then just do regular addition. I'm sure the rest of the adder is correct, I just do not know what is wrong with the part where I'm trying to convert.

Answer 1

You are using a lower case "y" in "Y = not(y) + 4'b0001;"

Also, you're using more additions than you need to. X-Y is the same thing as NOT(NOT(X)+Y).

Answer 2

put the if statement within an initial block
http://www.asic-world.com/verilog/vbehave1.html

Answer 3

reg [n-1:0] Y_compl;

always @( Z, Y, X, carryin ) begin
  Y_ = ( ~Y + 4'b0001 );
  if ( Z == 1'b0 ) begin
    {carryout, S} = X + Y_compl + carryin;
    overflow = carryout ^ X[n-1] ^ Y_compl[n-1] ^ S[n-1];
  end
  else begin
    {carryout, S} = X + Y + carryin;
    overflow = carryout ^ X[n-1] ^ Y[n-1] ^ S[n-1];
  end

end

A couple of important points.

1. Put the if statement inside the always block. Do not use two always blocks, you'll create a race condition in the simulator.
2. I created a new variable, Y_ because using Y, which is an input, remember, on the left hand side of an assignment will probably infer latches or do something else nasty when you synthesize.
3. I suggest using the bitwise inversion operator '~' to invert Y instead if the 'not' primitive. The synthesis tool has more freedom to optimize your code this way.
4. Double check for correct results, it's been awhile since I built an adder.