Verilog code to implement a 4-bit shift register circuit with a parallel load.

Here’s an example Verilog code to implement a 4-bit shift register circuit with a parallel load:

module shift_reg (output reg [3:0] out, input [3:0] in, input shift, input load);

    always @(*) begin
        if (load)
            out = in;
        else if (shift)
            out = {out[2:0], 1'b0};
    end

endmodule

This code defines a module called “shift_reg” that implements a 4-bit shift register with a parallel load. The output “out” is a registered output, which means that it is updated only on clock edges. The input “in” is a 4-bit vector that contains the data to be loaded into the shift register, and “shift” and “load” are control signals that determine whether the shift register should be shifted or loaded.

The “always @(*)” block is a combinational logic block that updates the “out” output based on the inputs and control signals. If “load” is high, then the “in” input is loaded into the shift register. Otherwise, if “shift” is high, then the shift register is shifted to the left by one bit, with the least significant bit replaced by a 0.

Note that this Verilog code assumes that the “out” output, the “in” input, and the “shift” and “load” control signals are all 1-bit signals. If you want to use multi-bit signals, you would need to modify the code accordingly. Additionally, this code implements a left shift register. If you wanted to implement a right shift register, you would need to modify the code accordingly.