Verilog has a ternary conditional operator ( ? : ) much like C:
(condition ? if_true : if_false)
This can be used to choose one of two values based on condition (a mux!) on one line, without using an if-then inside a combinational always block.
(0 ? 3 : 5) // This is 5 because the condition is false. (sel ? b : a) // A 2-to-1 multiplexer between a and b selected by sel. always @(posedge clk) // A T-flip-flop. q <= toggle ? ~q : q; always @(*) // State transition logic for a one-input FSM case (state) A: next = w ? B : A; B: next = w ? A : B; endcase assign out = ena ? q : 1'bz; // A tri-state buffer ((sel[1:0] == 2'h0) ? a : // A 3-to-1 mux (sel[1:0] == 2'h1) ? b : c )
A Bit of Practice
Given four unsigned numbers, find the minimum. Unsigned numbers can be compared with standard comparison operators (a < b). Use the conditional operator to make two-way min circuits, then compose a few of them to create a 4-way min circuit. You'll probably want some wire vectors for the intermediate results.
Expected solution length: Around 5 lines.
module top_module ( input [7:0] a, b, c, d, output [7:0] min);
Write your solution here