Question

What is the function implemented by the following Verilog modules:

Module FUNC1 (I0, I1, S, out):

input I0, I1;

input S;

output out;

out = S? I1: I0;

endmodule

module FUNC2 (out, ctl, clk,reset);

output [7:0] out;

input ctl, clk, reset;

reg [7:0] out;

always @(posedge clk)

if (reset) begin

out <= 8’b0;

end

else if (ctl) begin

out <= out+1;

end

else begin

out <= out-1;

end

endmodule

Short answer

The FUNC1 is the implementation of $2\times 1$multiplexer. The FUNC2 is the implementation of an 8-bit up-down counter.

Step-by-step solution

Analyzing the Verilog module FUNC1

In the module FUNC1, there are two inputs I0 and I1. There is another input S which acts as a selector for the inputs I0 and I1. The output variable is out. The module sets the value of the output using the ternary operator. If the value of S is 1, then the output is equal to I1. Otherwise, the output is I0. So, it is the implementation of $2\times 1$multiplexer which selected selects one value from the two inputs based on the selector.

Analyzing the Verilog module FUNC2

In module FUNC2, on the positive rising edge of the clock, the output is changed. If it is a reset, then all the eight bits are set to zero. If the up direction is asserted using ctl, then the output value is incremented by 1. Otherwise, it is decremented by 1. If the total bits used are 8, then the program is a counter 8-bit up-down counter.