69. Binary Subtractor

How do you plan to solve it?

simply bout is the not of carry generated

/*Write your code here*/
module sub4_2c(
    input [3:0] a,b,
    output [3:0] diff,
    output bout
);
    assign diff = a + ~b +4'd1;
    assign bout = (a[3]<b[3]) | (a[3]==b[3] && (a[2]<b[2])) | ((a[3]==b[3] && a[2]==b[2]) && (a[1]<b[1])) | ((a[3]==b[3] && a[2]==b[2] && a[1]==b[1]) && a[0]<b[0]);
endmodule

// simpler way, borrow is the not of carry generated during the sum
// wire c;
// assign {c,diff} = a + ~b + 4'd1;
// assign bout = ~c;