Create a 100-bit binary ripple-carry adder by instantiating 100 full adders. The adder adds two 100-bit numbers and a carry-in to produce a 100-bit sum and carry out. To encourage you to actually instantiate full adders, also output the carry-out from each full adder in the ripple-carry adder. cout[99] is the final carry-out from the last full adder, and is the carry-out you usually see.
module top_module(
input [99:0] a, b,
input cin,
output [99:0] cout,
output [99:0] sum );
genvar i;
generate
for(i=0;i<100;i++) begin:adder
if(i==0)
assign {cout[0],sum[0]} = a[0]+b[0]+cin;
else
assign {cout[i],sum[i]} = a[i]+b[i]+cout[i-1];
end
endgenerate
endmodule
这道题,先要把i==0这种情况剔除,再进入循环,因为cout[i-1]在i取0的时候是负数。
You are provided with a BCD one-digit adder named bcd_fadd that adds two BCD digits and carry-in, and produces a sum and carry-out.
module bcd_fadd ( input [3:0] a, input [3:0] b, input cin, output cout, output [3:0] sum );
Bcdadd100
Instantiate 100 copies of bcd_fadd to create a 100-digit BCD ripple-carry adder. Your adder should add two 100-digit BCD numbers (packed into 400-bit vectors) and a carry-in to produce a 100-digit sum and carry out.
module top_module(
input [399:0] a, b,
input cin,
output cout,
output [399:0] sum );
wire [99:0] cout_temp;
genvar i;
generate
for(i=0;i<100;i++) begin:bcd_fadd
if(i == 0)
bcd_fadd inst( a[3:0],
b[3:0],
cin,
cout_temp[0],
sum[3:0]
);
else
bcd_fadd inst( a[4*i+3:4*i],
b[4*i+3:4*i],
cout_temp[i-1],
cout_temp[i],
sum[4*i+3:4*i]
);
end
assign cout = cout_temp[99];
endgenerate
endmodule
需要考虑一个问题,前后一共100次例化连接,要不要取不同的名字。
module top_module (
input in,
output out);
assign out=in;
endmodule
module top_module (
output out);
assign out=0;
endmodule
module top_module (
input in1,
input in2,
output out);
assign out=~(in1|in2);
endmodule
module top_module (
input in1,
input in2,
output out);
assign out=in1&(~in2);
endmodule