[HDLBits] Module fadd

In this exercise, you will create a circuit with two levels of hierarchy. Your top_module will instantiate two copies of add16 (provided), each of which will instantiate 16 copies of add1 (which you must write). Thus, you must write two modules: top_module and add1.

Like module_add, you are given a module add16 that performs a 16-bit addition. You must instantiate two of them to create a 32-bit adder. One add16 module computes the lower 16 bits of the addition result, while the second add16 module computes the upper 16 bits of the result. Your 32-bit adder does not need to handle carry-in (assume 0) or carry-out (ignored).

Connect the add16 modules together as shown in the diagram below. The provided module add16 has the following declaration:

module add16 ( input[15:0] a, input[15:0] b, input cin, output[15:0] sum, output cout );

Within each add16, 16 full adders (module add1, not provided) are instantiated to actually perform the addition. You must write the full adder module that has the following declaration:

module add1 ( input a, input b, input cin, output sum, output cout );

Recall that a full adder computes the sum and carry-out of a+b+cin.

In summary, there are three modules in this design:

  • top_module — Your top-level module that contains two of...
  • add16, provided — A 16-bit adder module that is composed of 16 of...
  • add1 — A 1-bit full adder module.


If your submission is missing a module add1, you will get an error message that says Error (12006): Node instance "user_fadd[0].a1" instantiates undefined entity "add1".

[HDLBits] Module fadd_第1张图片

module top_module (
    input [31:0] a,
    input [31:0] b,
    output [31:0] sum
);//
	wire cout1,cout2;
	wire [15:0] re1,re2;
	add16 add_1(a[15:0],b[15:0],15'b0,re1,cout1);
	add16 add_2(a[31:16],b[31:16],cout1,re2,cout2);
	assign sum={re2,re1};
    
endmodule
	
module add1 ( input a, input b, input cin, output sum, output cout );
    assign {cout,sum}=a+b+cin;
// Full adder module here
	
endmodule

 这次全用的wire没用reg,因为发现上一题的reg是不必要的

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