题目:Module cseladd
One drawback of the ripple carry adder (See previous exercise) is that the delay for an adder to compute the carry out (from the carry-in, in the worst case) is fairly slow, and the second-stage adder cannot begin computing its carry-out until the first-stage adder has finished. This makes the adder slow. One improvement is a carry-select adder, shown below. The first-stage adder is the same as before, but we duplicate the second-stage adder, one assuming carry-in=0 and one assuming carry-in=1, then using a fast 2-to-1 multiplexer to select which result happened to be correct.
- ripple carry adder(见之前的练习)的一个缺点是,加法器计算carry out(从carry-in,在最坏的情况下)的延迟相当慢,而第二阶段的加法器在第一阶段的加法器完成之前无法开始计算其carry out。这使得加法器变慢。一种改进是carry-select adder,如下所示。第一阶段的加法器与之前相同,但我们复制了第二阶段的加法器,一个假设carry-in=0,另一个假设carry-in=1,然后使用快速的2-to-1多路复用器选择哪个结果是正确的。
In this exercise, you are provided with the same module add16 as the previous exercise, which adds two 16-bit numbers with carry-in and produces a carry-out and 16-bit sum. You must instantiate three of these to build the carry-select adder, using your own 16-bit 2-to-1 multiplexer.
- 在本练习中,您将使用与之前的练习相同的模块add16,该模块将两个16位数字与carry-in相加并产生一个carry-out和16位和。您必须实例化三个这样的模块来构建carry-select adder,使用您自己的16位2-to-1多路复用器。
Connect the modules together as shown in the diagram below. The provided module add16 has the following declaration:
- 将模块按照下面的图示连接起来。提供的模块add16具有以下声明:
module add16 ( input[15:0] a, input[15:0] b, input cin, output[15:0] sum, output cout );
在上一题Module fadd中,高16位的计算需要在得出进位信号后才能进行,这会使加法器变慢。同时由于进位信号只有0或1两种可能,所以在本题Module cseladd中,使用了两个add16来计算高16位的结果,并且这两个add16的进位信号cin分别为0和1。此时,在计算加法的计算步骤为,低16位相加(进位为0)、高16位相加(进位为0)与高16位加法(进位为1)三部分并行进行,再根据低16位相加得出的进位结果选取正确的高16位相加结果,进而得出最终答案。
objectivec
module top_module(
input [31:0] a,
input [31:0] b,
output [31:0] sum
);
//进位信号
wire cout;
//存储 高16位相加(进位为0) 的结果
wire [15:0] sum_high_1;
//存储 高16位相加(进位为0) 的结果
wire [15:0] sum_high_2;
//低16位相加
add16 add16_init_0(
.a(a[15:0]),
.b(b[15:0]),
.cin(1'b0),
.sum(sum[15:0]),
.cout(cout)
);
//高16位相加(进位为0)
add16 add16_init_1(
.a(a[31:16]),
.b(b[31:16]),
.cin(1'b0),
.sum(sum_high_1),
.cout()
);
//高16位相加(进位为1)
add16 add16_init_2(
.a(a[31:16]),
.b(b[31:16]),
.cin(1'b1),
.sum(sum_high_2),
.cout()
);
//根据进位结果选取正确的高16位运算结果
assign sum[31:16] = cout ? sum_high_2 : sum_high_1;
endmodule