HDLBits 系列（43）找 bug 专题

文章目录

例题1
- 原题复现
- 题目分析
- 改进之后
例题2
- 原题复现
- 题目分析
- 改进程序
例题3
- 原题复现
- 题目分析
- 改进程序
例题4
- 原题复现
- 题目分析
- 改进程序
例题5
- 原题复现
- 题目分析
- 改进程序

例题1

原题复现

原题链接
This 8-bit wide 2-to-1 multiplexer doesn’t work. Fix the bug(s).

module top_module (input sel,input [7:0] a,input [7:0] b,output out  );assign out = (~sel & a) | (sel & b);endmodule

题目分析

sel作为选择信号，1的话选择b输出，否则选择a输出。

首先第一个问题在于输出的位宽定义有问题；
由于a和b都是8位的信号，但是通过位运算符和sel进行运算，位宽不对应，需要扩展。

改进之后

module top_module (input sel,input [7:0] a,input [7:0] b,output [7:0] out  );assign out = ({8{sel}} & a) | (~{8{sel}} & b);endmodule

例题2

原题复现

原题链接
This three-input NAND gate doesn’t work. Fix the bug(s).

You must use the provided 5-input AND gate:

module andgate ( output out, input a, input b, input c, input d, input e );


module top_module (input a, input b, input c, output out);//andgate inst1 ( a, b, c, out );endmodule

题目分析

由于采用的例化方式是位置对应方式，所以，例化时候位置需要严格对应。

改进程序

module top_module (input a, input b, input c, output out);//wire out_mid;andgate inst1 ( out_mid, a, b, c,1, 1 );assign out = ~out_mid;endmodule

例题3

原题复现

原题链接
This 4-to-1 multiplexer doesn’t work. Fix the bug(s).

You are provided with a bug-free 2-to-1 multiplexer:

module mux2 (
input sel,
input [7:0] a,
input [7:0] b,
output [7:0] out
);


module top_module (input [1:0] sel,input [7:0] a,input [7:0] b,input [7:0] c,input [7:0] d,output [7:0] out  ); //wire mux0, mux1;mux2 mux0 ( sel[0],    a,    b, mux0 );mux2 mux1 ( sel[1],    c,    d, mux1 );mux2 mux2 ( sel[1], mux0, mux1,  out );endmodule

题目分析

问题1，在于中间变量的位宽定义不正确；
问题2，例化名不要和模块名一致；
问题3，第二个例化程序的选择变量有问题；

改进程序

module top_module (input [1:0] sel,input [7:0] a,input [7:0] b,input [7:0] c,input [7:0] d,output [7:0] out  ); //wire [7:0] mux0, mux1;mux2 inst_mux0 ( sel[0],    a,    b, mux0 );mux2 inst_mux1 ( sel[0],    c,    d, mux1 );mux2 inst_mux2 ( sel[1], mux0, mux1,  out );endmodule

例题4

原题复现

题目链接
The following adder-subtractor with zero flag doesn’t work. Fix the bug(s).

// synthesis verilog_input_version verilog_2001
module top_module ( input do_sub,input [7:0] a,input [7:0] b,output reg [7:0] out,output reg result_is_zero
);//always @(*) begincase (do_sub)0: out = a+b;1: out = a-b;endcaseif (~out)result_is_zero = 1;endendmodule

题目分析

一个always组合逻辑块内不要既有case又有其他的逻辑，有的允许，有的不允许，尽量不用。
out全为0时候，其自或运算为0，否则为1；

改进程序

// synthesis verilog_input_version verilog_2001
module top_module ( input do_sub,input [7:0] a,input [7:0] b,output reg [7:0] out,output  result_is_zero
);//always @(*) begincase (do_sub)0: out = a+b;1: out = a-b;endcaseendassign result_is_zero = ~(|out)?1:0;
endmodule

例题5

原题复现

原题链接

This combinational circuit is supposed to recognize 8-bit keyboard scancodes for keys 0 through 9. It should indicate whether one of the 10 cases were recognized (valid), and if so, which key was detected. Fix the bug(s).

module top_module (input [7:0] code,output reg [3:0] out,output reg valid=1 );//always @(*)case (code)8'h45: out = 0;8'h16: out = 1;8'h1e: out = 2;8'd26: out = 3;8'h25: out = 4;8'h2e: out = 5;8'h36: out = 6;8'h3d: out = 7;8'h3e: out = 8;6'h46: out = 9;default: valid = 0;endcaseendmodule

题目分析

为了把程序改错，真是用心恶毒呀，位宽改，进制改，逻辑改。。。好吧。还好可以实时调试。

改进程序

module top_module (input [7:0] code,output reg [3:0] out,output reg valid=1 );//always @(*) beginvalid = 0;out = 0;case (code)8'h45: beginout = 0;valid = 1;end8'h16: beginout = 1;valid = 1;end8'h1e: beginvalid = 1; out = 2;end8'h26: beginout = 3;valid = 1; end8'h25: beginout = 4;valid = 1;end8'h2e: beginvalid = 1;out = 5;end8'h36: beginvalid = 1;out = 6;end8'h3d: beginvalid = 1;out = 7;end8'h3e: beginvalid = 1;out = 8;end8'h46: beginvalid = 1;out = 9;endendcaseendendmodule