题解 | #位拆分与运算#
位拆分与运算
https://www.nowcoder.com/practice/1649582a755a4fabb9763d07e62a9752
1.题目
输入一个包含了四个数据[3:0][7:4][11:8][15:12] 的16位数据,
根据sel选择输出四个数据的相加结果,并输出valid_out信号(不输出时拉低)
0: 不输出且只有此时的输入有效 (即只有当sel=0时,输入的d才是有效的)
1:输出[3:0]+[7:4]
2:输出[3:0]+[11:8]
3:输出[3:0]+[15:12]
sel |
输出out[4:0] |
Valid_out |
00 |
不输出,并且此时的d有效 |
拉低 |
01 |
d[3:0]+d[7:4] |
拉高 |
10 |
d[3:0]+d[11:8] |
拉高 |
11 |
d[3:0]+d[15:12] |
拉高 |
3.参考波形图
4.解题思路
使用时序逻辑,在sel=0时,对输入d进行锁存,这样保证了在sel=0时更新d,sel≠0时保持锁存的值不变;sel=1/2/3时,通过使用锁存的d分别输出不同out和valid_out即可
5.绘制波形
initial
begin
rst = 1'b0;
d = 16'b0000_0000_0000_0000;
sel = 2'd0;
#30;
rst = 1'b1;
#10;
d = 16'b1000_0100_0010_0001;
#20;
sel = 2'd2;
#60;
sel = 2'd1;
d = 16'b1000_0100_0010_0011;
#60;
sel = 2'd0;
#20;
sel = 2'd3;
#20;
d = 16'b1000_0100_0010_0111;
#100;
$stop;
end
initial
begin
rst = 1'b0;
d = 16'b0000_0000_0000_0000;
sel = 2'd0;
#30;
rst = 1'b1;
#10;
d = 16'b1000_0100_0010_0001;
#20;
sel <= 2'd2;
#60;
sel = 2'd1;
d = 16'b1000_0100_0010_0011;
#60;
sel = 2'd0;
#20;
sel = 2'd3;
#20;
d = 16'b1000_0100_0010_0111;
#100;
$stop;
end