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我試圖實現一個verilog程序,大部分測試用例都通過了(1440中的1,188)。然而,我的問題是,我期望的溢出輸出當前顯示爲0,而期望值應該是1.Verilog測試平臺實現
繼承人打印到日誌的兩個示例,期望值不正確(一直滾動向右):
in1=1000000000000000 in2=1000000000000000 opCode=1001 result= 0111111111111111 expectedResult= 0111111111111111 overflow=0 expectedOverflow=1 in1=-32768 in2=-32768 opCode= 9 result= 32767 expectedResult= 32767 overflow=0 expectedOverflow=1
in1=1000000000000000 in2=1000000000000001 opCode=1001 result= 0111111111111111 expectedResult= 0111111111111111 overflow=0 expectedOverflow=1 in1=-32768 in2=-32767 opCode= 9 result= 32767 expectedResult= 32767 overflow=0 expectedOverflow=1
我找不到我確實出錯的地方。所以我想我的問題是,我究竟做錯了什麼?謝謝!
繼承人蔘考我的Verilog代碼的實現:
module Calculator(in1,in2,opCode,result,overflow);
input signed[15:0] in1, in2;
input[3:0] opCode;
output reg signed[15:0] result;
output reg overflow;
always @ (*) begin
if(opCode == 0000) begin
if(in1+in2<=32767 & in1+in2>= -32768) begin
overflow = 0;
end
else
begin
overflow = 1;
end
end
end
always @ (*) begin
if(opCode == 0001) begin
if(in1-in2<=32767 & in1-in2>= -32768) begin
overflow = 0;
end
else
begin
overflow = 1;
end
end
end
always @ (*) begin
if(opCode == 0010) begin
if(in1*5<=32767 & in1*5>= -32768) begin
overflow = 0;
end
else
begin
overflow = 1;
end
end
end
always @ (*) begin
if(opCode == 0011) begin
if ((in1 % 10) == 0) begin
overflow = 0;
end else begin
overflow = 1;
end
end
end
always @ (*) begin
if(opCode == 0100) begin
overflow = 0;
end
end
always @ (*) begin
if(opCode == 0101) begin
overflow = 0;
end
end
always @ (*) begin
if(opCode == 0110) begin
overflow = 0;
end
end
always @ (*) begin
if(opCode == 0111) begin
overflow = 0;
end
end
always @ (*) begin
if(opCode == 1000) begin
if(in1 == 32767) begin
overflow = 1;
end
else begin
overflow = 0;
end
end
end
always @ (*) begin
if(opCode == 1001) begin
if(in1==-32768) begin
overflow = 1;
end
else
begin
overflow = 0;
end
end
end
always @ (*) begin
case(opCode)
4'b0000: result = in1+in2; //add
4'b0001: result = in1-in2; //subtract
4'b0010: result = in1*5; //mult by 5
4'b0011: result = in1/10; //divide by 10
4'b0100: result = in1&in2; //AND
4'b0101: result = in1^in2; //XOR
4'b0110: result = in1|in2; //OR
4'b0111: result = /*((2^16)-1)-in1;*/(-(in1))-1; //complement
4'b1001: result = in1-1; //decrement
4'b1000: result = in1+1; //increment
endcase
end
endmodule
你的代碼不會合成,因爲'overflow'被分配在幾個總是block的地方。它只能被分配在一個總是阻止合成。 – Greg
'1001'是十進制一千和一個,'4'b1001'是二進制的十進制九 – Greg