I had to think a little bit, and here's what I came up with:
module addsub(
input [31:0] a,
input [31:0] b,
input Cin,
output [31:0] c,
output Cout,
input sub);
wire [31:0] b0 = {32{sub}} ^ b;
wire [33:0] b1 = {1'b0, b0, 1'b1};
wire [33:0] a1 = {1'b0, a, Cin};
wire [33:0] sum = b1 + a1;
assign Cout = sum[33];
assign c = sum[32:1];
endmodule
This computes the following function:
{Cout, c} = sub ? (a-b+!c} : {a+b+c}
You get a subtraction by taking the 1's complement of b and adding.
If Cin is zero, then you get a borrow, or a-b-1. If Cin is 1, then
that effectively turns the 1's complement into a 2's complement,
giving a straight subtraction.
Also, the way did it in the past yields a similar or maybe even
identical circuit, but only after the optimizer gets through wrangling
with it, and it's represented in a convoluted way. The above is much
better.
Thanks, Yann, for teaching me basic math. :)
--
Timothy Normand Miller
http://www.cse.ohio-state.edu/~millerti
Open Graphics Project
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