HDLBits刷题-Multiplexers

This is 目录

Mux2to
Mux2to1v
Mux9to1v
Mux256to1
Mux256to1v

Mux2to

Create a one-bit wide, 2-to-1 multiplexer. When sel=0, choose a. When sel=1, choose b.

二选一用一个条件语句即可解决

//成功代码
module top_module( input a, b, sel,output out ); assign out = (!sel)?a:b;
endmodule

波形

Mux2to1v

Create a 100-bit wide, 2-to-1 multiplexer. When sel=0, choose a. When sel=1, choose b.

只是把位宽扩展到100位。

//成功代码
module top_module( input [99:0] a, b,input sel,output [99:0] out );assign out = (sel)?b:a;
endmodule

波形

Mux9to1v

Create a 16-bit wide, 9-to-1 multiplexer. sel=0 chooses a, sel=1 chooses b, etc. For the unused cases (sel=9 to 15), set all output bits to ‘1’.

一个case-endcase语句可以解决

//成功代码
module top_module( input [15:0] a, b, c, d, e, f, g, h, i,input [3:0] sel,output [15:0] out );always @(*)begincase (sel)4'd0:out=a;4'd1:out=b;4'd2:out=c;4'd3:out=d;4'd4:out=e;4'd5:out=f;4'd6:out=g;4'd7:out=h;4'd8:out=i;default:out=16'hffff;endcaseend
endmodule

波形

Mux256to1

Create a 1-bit wide, 256-to-1 multiplexer. The 256 inputs are all packed into a single 256-bit input vector. sel=0 should select in[0], sel=1 selects bits in[1], sel=2 selects bits in[2], etc.

Hint:With this many options, a case statement isn’t so useful.
Vector indices can be variable, as long as the synthesizer can figure out that the width of the bits being selected is constant. In particular, selecting one bit out of a vector using a variable index will work.

根据提示可以知道：
矢量索引可以是可变的，只要合成器能够计算出所选位的宽度是恒定的。特别是使用可变索引从向量中选择一位将起作用。

//成功代码
module top_module( input [255:0] in,input [7:0] sel,output out );assign out = in[sel];
endmodule

Mux256to1v

Create a 4-bit wide, 256-to-1 multiplexer. The 256 4-bit inputs are all packed into a single 1024-bit input vector. sel=0 should select bits in[3:0], sel=1 selects bits in[7:4], sel=2 selects bits in[11:8], etc.

Expected solution length: Around 1–5 lines.

Hint:
With this many options, a case statement isn’t so useful.
Vector indices can be variable, as long as the synthesizer can figure out that the width of the bits being selected is constant. It’s not always good at this. An error saying “… is not a constant” means it couldn’t prove that the select width is constant. In particular, in[ sel4+3 : sel4 ] does not work.
Bit slicing (“Indexed vector part select”, since Verilog-2001) has an even more compact syntax.

//成功代码
module top_module( input [1023:0] in,input [7:0] sel,output [3:0] out );assign out = {in[4*sel+3], in[4*sel+2], in[4*sel+1], in[4*sel]};
endmodule

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HDLBits刷题-Multiplexers

This is 目录

Mux2to

Mux2to1v

Mux9to1v

Mux256to1

Mux256to1v

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