What is the difference between reg and wire in a verilog module? - verilog

When are we supposed to use reg and when are we supposed to use wire in a verilog module?
I have also noticed sometimes that a output is declared again as a reg. E.g reg Q in a D flip flop. I have read this somewhere - "The target output of procedural assignment statements must be of reg data type."
What are procedural assignment statements?
I have thoroughly googled this but was not able to find a clear explanation.

Wire:-
Wires are used for connecting different elements. They can be treated as physical wires. They can be read or assigned. No values get stored in them. They need to be driven by either continuous assign statement or from a port of a module.
Reg:-
Contrary to their name, regs don't necessarily correspond to physical registers. They represent data storage elements in Verilog/SystemVerilog. They retain their value till next value is assigned to them (not through assign statement). They can be synthesized to FF, latch or combinatorial circuit. (They might not be synthesizable !!!)
Wires and Regs are present from Verilog timeframe. SystemVerilog added a new data type called logic to them. So the next question is what is this logic data type and how it is different from our good old wire/reg.
Logic:-
As we have seen, reg data type is bit mis-leading in Verilog. System Verilog's logic data type addition is to remove the above confusion. The idea behind is having a new data type called logic which at least doesn't give an impression that it is hardware synthesizable. Logic data type doesn't permit multiple drivers. It has a last assignment wins behavior in case of multiple assignments (which implies it has no hardware equivalence). Reg/Wire data types give X if multiple drivers try to drive them with different values. Logic data type simply assigns the last assignment value. The next difference between reg/wire and logic is that logic can be both driven by assign block, output of a port and inside a procedural block like this
logic a;
assign a = b ^ c; // wire style
always (c or d) a = c + d; // reg style
MyModule module(.out(a), .in(xyz)); // wire style

Procedural blocks refers to always, always_ff, always_comb, always_latch, initial etc. blocks. While procedural assignment statements refers to assigning values to reg, integer etc., but not wires(nets).
wire elements must be continuously driven by something, and cannot store a value. Henceforth, they are assigned values using continuous assignment statements.
reg can be used to create registers in procedural blocks. Thus, it can store some value.
reg elements can be used as output within an actual module declaration. But,reg elements cannot be connected to the output port of a module instantiation.
Thus, a reg can drive a wire as RHS of an assign statement. On the other way round, a wire can drive a reg in as RHS of a procedural block.
For clear idea about declaration of reg or wire, refer the image below:
So, whenever inferring to sequential logic, which stores/holds some value, declare that variable/port as reg. Here, Q is a reg inside a module, but while instantiating this module inside some other module, then this port must be connected to a wire.
Remember, wire can only infer to combinational logic, while reg can infer to either combinational or sequential logic.
Dave's blog is a good source for detailed information. For further information, refer to synthesizing difference and Verilog wire-reg links.

Simple difference between reg and wire is, the reg is used in combinational or sequential circuit in verilog and wire is used in combinational circuit
reg is used to store a value but wire is continuely driven some thing and wire is connected to outport when module initialization but reg is con not connected

Related

What does the variable name in the register declaration indicate (Verilog)

I'm a just starting to learn verilog , and I'm having trouble with some things. I've found a few resources to help, but there are some things which aren't clear and I need specified. I have code for a D Flip-flop below. And I understand how the declaration for in and out work. What I dont get is the register.
When it says out is the variable associated with it.Does this association mean, out is the register along with output? Or does it mean out is also the output for the register?
module DFF(quarter, in, out) ;
parameter n = 1; // width
input quarter ;
input [n-1:0] in ;
output [n-1:0] out ;
reg [n-1:0] out ;
always #(quarter=1)
out = in ;
endmodule
In Verilog there are nets and there are variables. The most common kind of net by far is a wire, which you're probably familiar with. You can declare a variable using var, but most people say reg, because that is how it always has been done.
(In Verilog, but not in SystemVerilog), wires have to be driven by
assign statements
the outputs of instantiated modules
and variables have to be driven from
initial and always blocks.
Your output out is driven from an always block and so must be a variable. The line
reg [n-1:0] out ;
declares the output out as being a variable (rather than a wire).
In fact, you are using an old-fashioned way of specifying inputs and outputs. Since 2001, the way to do it is like this:
module DFF #(parameter n = 1)
(input wire quarter,
input wire [n-1:0] in,
output reg [n-1:0] out);
always #(quarter=1)
out = in ;
endmodule
This is the so-called ANSI form and, I think you would agree, is more sensible. I would advise using this form, not the old-fashioned form you are using.
BTW, you code is a bit strange. I'm not entirely sure what it's supposed to do, but this if you're expecting output to be driven by in when quarter is 1, I'd do it more like this:
module DFF // #(parameter n = 1) // you're not using parameter n
(input wire quarter,
input wire [n-1:0] in,
output reg [n-1:0] out);
always #(*)
if (quarter == 1'b1)
out = in ;
else
// IMPORTANT ! : what is out when quarter != 1 ?
// your current code would synthesise to a latch
// is that what you wanted? (And if so, are you
// sure that's what you wanted)
endmodule
initially verilog was though as a language which does behavioral types of operation on state variables (registers, reg) and connects them via nets. So, reg is a variable which is supposed to keep its value between operations, net is just a connection and has no state associated with it, i.e. wire. reg values can be calculated in procedural blocks, i.e. always block, wires can be connected only (using continuous assignments, assign statements (outside of the always blocks)).
There are different rules for using for regs and wires and different behavior of them in simulation, in particular around multiple drivers. See Matthew's reply for more info.

Verilog multiplication through repeated addition

I need help with a Verilog design I'm doing.
the idea is to do multiplication through repeated addition every time the M bit is set to 1/true. then I need to output that value. The assignment statement F=P; is throwing the error.
This is the error I'm getting
Error (10044): Verilog HDL error at Design2.v(13): expression cannot reference entire array "P"
It is my understanding that I can assign a register to an output if they are the same size. If I'm wrong then how can I implement this?
module Design2(A, N, M, F);
input A[7:0];
input N[3:0];
input M;
reg P[15:0];
output F[15:0];
always #(M) begin
repeat(N) begin
P = P + A;
end
F=P;
end
endmodule
Keeping apart the logic side, I can see two issues with your design.
As the comments pointed out, you have used unpacked arrays at wrong place.
Declaration of output port F.
For the first issue, the design needs a 16-bit vector for all the ports and variables. Here, the design takes a bunch of 8-bits as input A which must be declared as input [7:0] A, which is a packed array. Also, P needs to be a vector of size 16-bit in order to have contiguous operations.
This is to be done so that, when the statement P=P+A executes, the addition operation is done with all the variables taken with their respective sizes (size if P=16 and A=8). The overall addition is to be done with 16-bits, padding zeros in MSB side of A.
Hence, convert all the variables to packed array as follows:
input [7:0] A;
input [3:0] N;
input M;
reg [15:0] P;
output reg [15:0] F;
For the second issue, the design assigns output in a procedural always block.
wire elements must be continuously driven by something, and cannot store a value. Henceforth, they are assigned values using continuous assignment statements.
reg can be used to create registers in procedural blocks. Thus, it can store some value.
When an input port is connected, it has to be driven through continuous assignments from parent module, hence input ports are always wire. While output from a module can be driven continuously through a wire, or can be driven through procedural assignments through reg. Following image shows the port connection rules.
To accomplish this in current design, the output port must be declared as output reg [15:0] F. This is the cause of compilation error shown.
Either way, you can have output [15:0] F with no reg and make continuous assignment to F as follows. This will synthesize F to wire:
output [15:0] F;
assign F=P;
SystemVerilog adds a logic datatype to remove confusion between usage of wire and reg declaration. A logic can be driven by both continuous assignment or blocking/non blocking assignment.
// Either continuous assignment
output logic [15:0] F;
assign F=P;
// Or procedural assignment
output logic [15:0] F;
// Inside always block
F=P;
For more information on packed and unpacked array, refer SystemVerilog Arrays link. Regarding port declarations, refer Wire and Net pdf. Refer SystemVerilog IEEE 1800-2012 for more information on logic datatype.

Verilog wire is not assigned

I was wondering that if a wire is declared in a Verilog code, but it is not assigned any value, does Verilog treat its value as ZERO ?
For example, I see a code where:
wire start;
module_if my_module_if(.clk(in_clk), .start(start));
Can I assume that the value of "start" will be zero?
Is this an acceptable style for Verilog?
Nets (including wire) without an assignment will be initialized to 'z' (aka high-impedance).
But if you are trying to use "start" as some kind variable or state, then you should probably declare it as "reg" or "logic" (System-Verilog) rather than "wire" (which tends to be used more for interconnect).
Style-wise, it may be less error-prone for you and for the reader of the code to be explicit about the initialization of your logic. For example ...
logic start;
initial begin
start = 1'b0;
end

How to declare an array of 4 bits in Verilog

I need an array to hold a 4-bit number. It is neither an input or an output, just an intermediary value necessary for calculations. How would this be declared?
You mean that you need a variable?
reg [3:0] mynumber;
For synthesis, you would use either a register or a wire, depending on what you needed it for.
reg [3:0] my_reg;
or
wire [3:0] my_wire;
If you will use this value in an always block, you need to declare it as a reg.
If you are using it for combinational logic, not inside an always block, you would declare it as a wire. This would be used with an assign statements or in a port list.

Why is there no assign statement in this Verilog function?

Coming from a C++ background I'm starting to learn Verilog. This code describes four inputs going into two AND gates. The outputs from those two AND gates go into an OR gate. The output from the OR gate is the final output.
// a user-defined AND gate
module my_and2 (in, out);
input [1:0] in;
output out;
assign out = in[1]&in[0];
endmodule
// a user-defined OR gate
module my_or2 (in, out);
input [1:0] in;
output out;
assign out = in[1]|in[0];
endmodule
// the AND-OR logic built on top of the user-defined AND and OR gates
module and_or (in_top, out_top);
input [3:0] in_top;
output out_top;
wire [1:0] sig;
// instantiate the gate-level modules
my_and2 U1 (.in(in_top[3:2]),.out(sig[1]));
my_and2 U2 (.in(in_top[1:0]),.out(sig[0]));
my_or2 U3 (.in(sig),.out(out_top));
endmodule
The first two modules make sense to me. However, the last one doesn't. The first two modules have an assign statement at the end to set the value for the output variable. However, the last one doesn't. Why is that?
Verilog is 'event driven'. When writing verilog, think in terms of sensitivity lists.
In your example of the AND gate, you've the expression assign out = in[1]&in[0];. Your expression is said to be sensitive to in[0] and in[1]. This means that any time in[0] or in[1] change, the expression will be recomputed, and the value of out will be updated.
So in your toplevel module and_or, you're basically building a big tree of expressions that are sensitive to the outputs of the preceding expressions. This tree is, of course, built using the module connections. So a change in the value of one of the inputs to this toplevel module will ripple through all expressions in its 'logic cone'.
To drive the inputs you'll need higher level testbench module driving signals into your and_or module. This will supply inputs spaced out in time which will trigger the expressions in and below and_or. If not, your sim will have no events, so no expressions will trigger and the sim will time-out at 0ps because it is 'event starved'.
PS: for your AND gate expression, assign out = ∈ will work too... (reduction AND operator)
out_top is driven by the U3 instance output.
To put things simply, I like to think instantiation as just connecting wires.
Modules are blocks of digital circuits. You AND and OR gate modules are where magic happens. You already understand that part. By instantiating those modules, it's like you're connecting the input wires of your top level module with inputs of two blocks AND module. Then taking the outputs of them and taping them to the input wire sticking out of your OR block. And finally you're connecting the output of OR block to the output signal wire of top level.

Resources