Verilog Is Not a Legal Net Lvalue

It confused me for a while at first. I mainly write struturalverilog (modules and continuous assignment), but FFs only work in behavioral verilog. vif.data <= trans.data; | xmelab: *E,WANOTL (. /verif/sram_driver.sv,36|10): A network is not a legal value in this context [9.3.1(IEEE)]. I basically set different control signals for the ALU to perform operations in Verilog. But I tried every possible way to write what I wanted, but in vain, you can help me. How should I set these control signals to specific 3-bit aluminum states? A network is not a legal value in this context" for all assign statements in the given code. You cannot insert an Assign statement into a procedural block. In Verilog, you can model combination circuits using assign or always. You can`t mix them together. This is my code, I have all kinds of assignments; (Sorry, it`s reversed and I don`t know how to rotate it because the website where I upload images automatically rotated it that way) I find that a simplified view of Verilog as a simulation language helps me understand the rules better. If you think like a simulator, Areg means that the simulator needs memory to calculate the current value. A cable does not require storage because it is continuously assigned or connected to another value for which the simulator already has memory.

This has nothing to do with the type of procedural block in which the reg or integer is assigned. For example, a “reg” does not necessarily imply storage in actual hardware. However, if you were to translate the Verilog procedure into a C program, the reg would require a variable. Are you getting a syntax error? Unexpected behavior? One more thing? Hi, I`m trying to put the output wire inside a block that is still procedural, but I get an “invalid l-value error”. I compile with iVerilog. Any help is greatly appreciated. >>reg [10:0] InKbWd;>>mod_kb_read _mod_kb_read(ExtRst,ExtKbClk,ExtKbData,InKbWd); >>endmodule>>module mod_kb_read (Rst input,> KbClk input,> KbData input,> reg [10:0] KbWd output); >Module _kb_echo (Yada Yada); Module mod_kb_read (Rst input, KbClk input, KbData input, reg [10:0] KbWd output); always @ (negedge KbClk or negedge Rst) begin KbWd[0] <= KbData; KbWd[1] <= KbWd[0]; KbWd[2] <= KbWd[1]; KbWd[3] <= KbWd[2]; KbWd[4] <= KbWd[3]; KbWd[5] <= KbWd[4]; KbWd[6] <= KbWd[5]; KbWd[7] <= KbWd[6]; KbWd[8] <= KbWd[7]; KbWd[9] <= KbWd[8]; KbWd[10] Thanks guys. I need to re-read the section on wire/net and reg/int.> These assignment rules confused me This rule, which seems a bit strange at first glance but actually makes a lot of sense, is very relaxed in SystemVerilog and therefore your original code should be accepted by your tools when you switch the necessary command-line switches. However, some of the low-end tools and FPGAs haven`t caught up with that yet, so it`s still a good plan to follow traditional verilog rules. – If a thing gets its value by procedural assignment, then that thing must be a variable (reg, integer or one of the other types).- If a thing gets its value in another way – especially by being driven by something on the other side of a port boundary or by a continuous assignment – then the thing must be a network (wire).

Please use code tags to make your code easier to read. I added them for you. The InKbWd signal must be declared as a (net) wire and not as a (variable) reg. “Continuous mapping” refers to the Assign statement that appears at the top level of a module: assign some_net = some_expression; Not to be confused with continuous procedural assignment – the assign keyword that appears in procedural code – which you should avoid like the plague.– Jonathan Bromley **************mod_kb_read _mod_kb_read******************<—****** I have a dilemma as to how to get the data at the DUT and what happens when the reading occurs. mod_kb_read _mod_kb_read(ExtRst,ExtKbClk,ExtKbData,InKbWd); The way I think about it is that reg retains its value, like a capacitor (in the era of dynamic logic) or a guardian (now). This allows the behavior mapping to work so that the value remains between assignments. A wire must always have a driver (unless it allows tristate). In fact, this is a perfectly acceptable, albeit sparse, message if you mistakenly connect module output to a registry. I`m a beginner at Verilog and had a hard time defining an if-else loop. The error message is: You should read this article and see if you can determine what the problem is. The full program can be found below.

All modules have been defined correctly and I am sure the fault is only in this part.