Verilog code to implement a 4-bit shift register circuit with a serial in/parallel out.

Here’s an example Verilog code to implement a 4-bit shift register circuit with a serial in/parallel out:

module shift_register (output reg [3:0] parallel_out, input serial_in, input shift);

    reg [3:0] shift_reg;

    always @(posedge shift) begin
        if (shift) begin
            shift_reg <= {shift_reg[2:0], serial_in};
        end
    end

    assign parallel_out = shift_reg;

endmodule

This code defines a module called "shift_register" that implements a 4-bit shift register with a serial in/parallel out. The output "parallel_out" is a registered output that represents the parallel output of the shift register. The input "serial_in" is the serial input to the shift register, and "shift" is the clock input that controls the shift operation.

The "reg [3:0] shift_reg;" statement declares a 4-bit register called "shift_reg" that stores the current state of the shift register.

The "always @(posedge shift)" block is a sequential logic block that updates the "shift_reg" register based on the clock and shift signals. If the "shift" signal is high, then the shift register shifts the current contents of "shift_reg" left by one bit and inserts the "serial_in" input at the least significant bit.

The "assign parallel_out = shift_reg;" statement is a combinational logic statement that assigns the current state of "shift_reg" to the "parallel_out" output.

Note that this Verilog code assumes that the "parallel_out" output, the "serial_in" and "shift" inputs 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 rising-edge triggered shift register. If you wanted to implement a falling-edge triggered shift register, you would need to modify the code accordingly.