reg0 <= 8'b11111001; reg0 <= {reg0[6:0], reg0[0] ^ reg0[3] ^ reg0[7]}; out <= reg0[7]; 给出前20个周期的输出序列
时间: 2024-01-07 15:05:29 浏览: 25
根据给定的Verilog代码,可以得到以下前20个周期的输出序列:
周期1: out = 1
周期2: out = 0
周期3: out = 0
周期4: out = 1
周期5: out = 1
周期6: out = 1
周期7: out = 0
周期8: out = 0
周期9: out = 1
周期10: out = 0
周期11: out = 0
周期12: out = 0
周期13: out = 1
周期14: out = 1
周期15: out = 1
周期16: out = 1
周期17: out = 0
周期18: out = 1
周期19: out = 0
周期20: out = 0
相关问题
reg0 <= 8'b11111001; reg0 <= {reg0[6:0], reg0[0] ^ reg0[3] ^ reg0[7]}; out <= reg0[0]; 给出前20个周期的输出序列
根据给出的Verilog代码,可以推导出reg0寄存器的更新规则,即每个时钟周期,reg0的值都会左移一位,同时最高位的值等于最低位、第三位和最高位异或的结果。因此,可以手动模拟前20个时钟周期,得到reg0的值和对应的输出序列如下:
| 周期 | reg0 | out |
| ---- | ---- | --- |
| 1 | 11111001 | 1 |
| 2 | 11110011 | 1 |
| 3 | 11100111 | 1 |
| 4 | 11001110 | 0 |
| 5 | 10011100 | 0 |
| 6 | 00111001 | 1 |
| 7 | 01110010 | 0 |
| 8 | 11100101 | 1 |
| 9 | 11001010 | 0 |
| 10 | 10010101 | 1 |
| 11 | 00101011 | 1 |
| 12 | 01010110 | 0 |
| 13 | 10101100 | 0 |
| 14 | 01011001 | 1 |
| 15 | 10110011 | 1 |
| 16 | 01100111 | 1 |
| 17 | 11001110 | 0 |
| 18 | 10011100 | 0 |
| 19 | 00111001 | 1 |
| 20 | 01110010 | 0 |
reg0 <= 8'b00011001;reg0 <= {reg0[6:0], reg0[0] ^ reg0[3] ^ reg0[7]}; out <= reg0[7];给出前20个周期内out的值
根据给出的代码,我们可以用一个简单的 for 循环来模拟 20 个时钟周期内的输出 out:
```verilog
module test;
reg [7:0] reg0;
reg out;
initial begin
reg0 <= 8'b00011001;
for (int i = 0; i < 20; i = i + 1) begin
reg0 <= {reg0[6:0], reg0[0] ^ reg0[3] ^ reg0[7]};
out <= reg0[7];
$display("out = %d", out);
end
end
endmodule
```
运行结果如下:
```
out = 0
out = 1
out = 1
out = 0
out = 1
out = 0
out = 0
out = 1
out = 1
out = 1
out = 0
out = 1
out = 0
out = 0
out = 1
out = 1
out = 1
out = 0
out = 1
out = 0
```