HDLBits 系列（23）3 输入的 LUT

目录

原题复现

审题

我的设计

---

原题复现

In this question, you will design a circuit for an 8x1 memory, where writing to the memory is accomplished by shifting-in bits, and reading is "random access", as in a typical RAM. You will then use the circuit to realize a 3-input logic function.

First, create an 8-bit shift register with 8 D-type flip-flops. Label the flip-flop outputs from Q[0]...Q[7]. The shift register input should be called S, which feeds the input of Q[0] (MSB is shifted in first). The enable input controls whether to shift. Then, extend the circuit to have 3 additional inputs A,B,C and an output Z. The circuit's behaviour should be as follows: when ABC is 000, Z=Q[0], when ABC is 001, Z=Q[1], and so on. Your circuit should contain ONLY the 8-bit shift register, and multiplexers. (Aside: this circuit is called a 3-input look-up-table (LUT)).

Module Declaration

module top_module (
    input clk,
    input enable,
    input S,
    input A, B, C,
    output Z ); 

---

审题

本题名字叫实现一个3输入的LUT，而所谓的LUT，可以看成是一个存储器，一张真值表，它列出了所有的输入对应的输出，通过给定输入，就可以得到输出。

本题的意思是首先设计一个移位寄存器，之后，通过输入ABC来选择输出。

我的设计

给出我的设计：

module top_module (
    input clk,
    input enable,
    input S,
    input A, B, C,
    output reg Z ); 
    
    reg[7:0] q;
    always@(posedge clk)begin
        if(enable)begin
            q<=	{q[6:0],S};
        end
    end
    
    wire[2:0] output_index = {A,B,C};
    always@(*) begin
        case(output_index)
            3'd0:Z=q[0];
            3'd1:Z=q[1];
            3'd2:Z=q[2];
            3'd3:Z=q[3];
            3'd4:Z=q[4];
            3'd5:Z=q[5];
            3'd6:Z=q[6];
            3'd7:Z=q[7];
            endcase
    end
    

endmodule