Verilog error: not a valid l-value

SasaS picture SasaS · Nov 15, 2017 · Viewed 9.2k times · Source

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:

Errors

uut is my instantiation unit in my testbench called main

Answer

Serge picture Serge · Nov 15, 2017

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