我正在编写一个Verilog代码,用8全加器构造一个8位加法器。该8位加法器应在8位总线中每个增加2个输入。以下是单全加器的代码:
module FullAdder(
input a_,
input b_,
input cin_,
output sout_,
output cout_
);
wire temp1, temp2, temp3;
assign sout_ = a_ ^ b_ ^ cin_;
assign cout_ = ((a_ & b_) | (b_ & cin_) | (cin_ & a_));
endmodule这是8位加法器护城河的代码,它将调用全加器8次。
module EightBitAdder(
input [7:0] a,
input [7:0] b,
//input cin,
output cout, //carry; will be sent as OP, but won't be further used.
output [7:0] sout //sum, sent as OP
);
wire try;
begin
FullAdder mg0(.a_(a[0]), .b_(b[0]), .cin_(0), .cout_(try), .sout_(sout[0]));
FullAdder mg1(.a_(a[1]), .b_(b[1]), .cin_(try), .cout_(try), .sout_(sout[1]));
FullAdder mg2(.a_(a[2]), .b_(b[2]), .cin_(try), .cout_(try), .sout_(sout[2]));
FullAdder mg3(.a_(a[3]), .b_(b[3]), .cin_(try), .cout_(try), .sout_(sout[3]));
FullAdder mg4(.a_(a[4]), .b_(b[4]), .cin_(try), .cout_(try), .sout_(sout[4]));
FullAdder mg5(.a_(a[5]), .b_(b[5]), .cin_(try), .cout_(try), .sout_(sout[5]));
FullAdder mg6(.a_(a[6]), .b_(b[6]), .cin_(try), .cout_(try), .sout_(sout[6]));
FullAdder mg7(.a_(a[7]), .b_(b[7]), .cin_(try), .cout_(try), .sout_(sout[7]));
end
endmodule问题是输出没有正确显示。它总是显示第一个位,然后用“不关心”(X)填充其余的位。这里有什么问题吗?
发布于 2016-11-06 13:11:30
相同的try线是由所有8个满加法器驱动的,所以这很可能会产生一个X值,如果两者都由0和1驱动,则由全加器执行。
考虑制作一条执行电线,例如:
wire [7:0] cout;并为不同的满加法使用不同的位元,例如:
FullAdder mg0(.a_(a[0]), .b_(b[0]), .cin_(0), .cout_(cout[0]), .sout_(sout[0]));
FullAdder mg1(.a_(a[1]), .b_(b[1]), .cin_(cout[0]), .cout_(cout[1]), .sout_(sout[1]));
...https://stackoverflow.com/questions/40447997
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