Verilog error: not a valid l-value

Question

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;                                


        

Answer 1

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

Answer 2

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