我正在为分配编写Verilog程序，要求为5位加减法器编写程序。运行该程序后，由于某种原因，我的输出为“ x”，因为我不熟悉Verilog。
module TestMod;
parameter STDIN = 32'h8000_0000; // I/O address of keyboard input channel
reg [7:0] str [1:3], str1[1:2];
reg [4:0] X, Y; // 5-bit X, Y to sum
reg Op;
wire [4:0] S; // 5-bit Sum to see as result
wire C4,C5; // like to know this as well from result of adder
initial
begin
$display("Enter X (two digit 00~15 since max is 01111): ");
str[1] = $fgetc(STDIN);
str[2] = $fgetc(STDIN);
str[3] = $fgetc(STDIN);
X = (str[1] - 48)*10 + (str[2] - 48);
$display("Enter Y (two digit 00~15 since max is 01111): ");
str[1] = $fgetc(STDIN);
str[2] = $fgetc(STDIN);
str[3] = $fgetc(STDIN);
Y = (str[1] - 48)*10 + (str[2] - 48);
$display("Enter either '+' or '-': ");
str1[1] = $fgetc(STDIN);
str1[2] = $fgetc(STDIN);
if(str1[1] == 43)
Op = str1[1] + 5;
else
Op = str1[1] + 4;
#20; // wait until adder gets them processed
$display("X = %d (%b)\tY = %d (%b)\tc0 = %d", X,X,Y,Y,Op);
$display("Result = %d (as unsigned %b)", S, S);
$display("C4 = %b\tC5 = %b\tE = %b",C4,C5,C4^C5);
if(C4^C5 == 0'b0)
$display("Since E is 0, C5 is not needed");
else
$display("Since E is 1, correct with C5 in front: %b%b", C4^C5, S);
end
BigAdder BA(X,Y,Op,S,C4,C5);
endmodule
module BigAdder(X, Y, Op, S, C4, C);
input [4:0] X, Y;
input Op;
output [4:0] S; // S should be similar
output C4,C; // another output for a different size
wire C1,C2,C0,C3,C5,Y1,Y2,Y3,Y4;
xor(Y1, Y[0], Op);
xor(Y2, Y[1], Op);
xor(Y3, Y[2], Op);
xor(Y4, Y[3], Op);
xor(C, C5, Op);
FullAdderMod a0(X[0], Y1, Op, S[0], C0);
FullAdderMod a1(X[1], Y2, C0, S[1], C1);
FullAdderMod a2(X[2], Y3, C1, S[2], C2);
FullAdderMod a3(X[3], Y4, C2, S[3], C3);
FullAdderMod a4(X[4], Y5, C3, S[4], C5);
endmodule
module FullAdderMod(X, Y, Cin, S, Cout); // single-bit adder module
input X,Y,Cin;
output Cout,S;
assign S = X ^ Y ^ Cin;
assign Cout = (X & Cin) | (X & Y) | (Y & Cin);
endmodule
在我的作业中，我必须按照教授的说明显示输出，如下所示：
atoz[35]% demo-a.out
Enter X (range 00 ~ 15):
01
Enter Y (range 00 ~ 15):
15
Enter either '+' or '-':
-
X=00001 ( 1) Y=01111 (15) C0=1
Result=10010 (as unsigned 18)
C4=0 C5=0 E=0
Since E is 0, C5 is not needed.
atoz[36]% demo-a.out
Enter X (range 00 ~ 15):
11
Enter either Y (range 00 ~ 15):
15
Enter either '+' or '-':
+
X=01011 (11) Y=01111 (15) C0=0
Result=11010 (as unsigned 26)
C4=1 C5=0 E=1
Since E is 1, correct with C5 in front: 011010
但是我的实际输出是：
ent_3$ ./a.out
Enter X (two digit 00~15 since max is 01111):
12
Enter Y (two digit 00~15 since max is 01111):
13
Enter either '+' or '-':
+
X = 12 (01100) Y = 13 (01101) c0 = 0
Result = X (as unsigned x1001)
C4 = z C5 = x E = x
Since E is 1, correct with C5 in front: xx1001
任何人都可以帮助我吗？