Verilog error: not a valid l-value

落花浮王杯 提交于 2021-02-05 10:31:39

问题


I'm trying to test if a wire(s) is on or not to signify if there is an error/overflow in my alu code. Given this code:

output reg[3:0]x;                       // line 149
output wire error;
output wire overflow;

always @* begin
    if(error || overflow) begin         
        assign x = 4'b1111;             // line 155
        assign error = ~error;
        assign overflow = ~overflow;
    end else begin
        assign x = opcode;
    end
end

I get following error messages:

uut is my instantiation unit in my testbench called main


回答1:


The code in the example has several issues.

1) you tried to use 'procedural assignments' which is an advanced verilog topic. In other words assign statement inside of an always block. This is not synthesizable, can only be used on reg types, and is there in verilog for very special cases. Do not use it.

You error messages coming from the fact that error and overflow are declared as wire.

2) you are trying to assign inverted version of a value to itself in a non-clocked logic. It will not behave the way you expect. Depending on usage it can either not toggle or will cause an infinite zero-delay loop, or in your case it could just generate a glitch.

So, potentially, your code should look something like the following:

input wire clk; // << you need clock
output reg[3:0]x;                       // line 149
output wire error;
output wire overflow;

reg error_reg, overflow_reg; 

 always @(posedge clk) begin
    if(error || overflow) begin         
        x <= 4'b1111;             // line 155
        error_reg <= ~error;
        overflow_reg <= ~overflow;
    end else begin
        x <= opcode;
    end
 assign error = error_reg;
 assign overflow = overflow_reg;
end



回答2:


Your using the assign incorrectly. That can be used outside of a always process, but not inside of one.
Also, the type wire, is required for an assign

wire [3:0] x;
assign x = 4'b1111;

Inside the always process, remove the assign statement and just say

reg [3:0] x;  // Note that this is assigned as a reg now
 always @* begin
    if(blah) begin 
       x = 4'b1111;
    end else begin
       x = opcode;
    end
 end


来源:https://stackoverflow.com/questions/47297001/verilog-error-not-a-valid-l-value

标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!