Verilog code to implement a 4-bit barrel shifter circuit.

Here’s an example Verilog code to implement a 4-bit barrel shifter circuit:

module barrel_shifter (output reg [3:0] shifted, input [3:0] data, input [1:0] shift_amt, input shift_dir);

    always @(*) begin
        case ({shift_dir, shift_amt})
            2'b00: shifted = data; // no shift
            2'b01: shifted = {data[2:0], 1'b0}; // shift left
            2'b10: shifted = {1'b0, data[3:1]}; // shift right
            2'b11: shifted = {data[1:0], data[3:2]}; // rotate left
        endcase
    end

endmodule

This code defines a module called “barrel_shifter” that implements a 4-bit barrel shifter. The output “shifted” is a registered output that represents the shifted data. The input “data” is the 4-bit input data to be shifted. The input “shift_amt” is a 2-bit input that represents the amount of shift. The input “shift_dir” is a 1-bit input that specifies the direction of the shift.

The “case” statement is a combinational logic block that calculates the shifted output based on the input signals. If “shift_dir” is 0, the output is the same as the input (“no shift”). If “shift_dir” is 1 and the shift amount is positive, the output is the input shifted left by the specified amount. If “shift_dir” is 2 and the shift amount is positive, the output is the input shifted right by the specified amount. If “shift_dir” is 3 and the shift amount is positive, the output is the input rotated left by the specified amount.

Note that this Verilog code assumes that the “shifted” output, the “data” input, and the “shift_amt” input are all 4-bit signals, and the “shift_dir” input is a 1-bit signal. If you want to use different signal widths, you would need to modify the code accordingly. Additionally, this code implements a combinational circuit, which means that the output is updated immediately whenever there is a change in the input signals. If you wanted to implement a registered circuit, you would need to modify the code accordingly.