I am trying to debug my code shown below. I am fairly new to SystemVerilog and hopefully I can learn from this. Let me know of any suggestions.
The errors I am receiving are:
Error-[ICPD] Invalid procedural driver combination
"divide.v", 2
Variable "Q" is driven by an invalid combination of procedural drivers.
Variables written on left-hand of "always_comb" cannot be written to by any
other processes, including other "always_comb" processes.
"divide.v", 2: logic [7:0] Q;
"divide.v", 8: always_comb begin
if (x <= R) begin
...
"divide.v", 5: Q = 8'b0;
Error-[ICPD] Invalid procedural driver combination
"divide.v", 2
Variable "R" is driven by an invalid combination of procedural drivers.
Variables written on left-hand of "always_comb" cannot be written to by any
other processes, including other "always_comb" processes.
"divide.v", 2: logic [7:0] R;
"divide.v", 8: always_comb begin
if (x <= R) begin
...
"divide.v",6: R = y;
My SystemVerilog Code is:
module divider(input logic [7:0] x,y,
output logic [7:0] Q,R);
initial
begin
Q = 8'd0;
R = y;
end
always_comb
begin
if (x<=R)
begin R <= R - x; Q <= Q + 8'd1; end
end
endmodule
module test1;
logic [7:0] x,y,Q,R;
divider Divider1 (x,y,Q,R);
initial
begin
x = 8'd2;
y = 8'd8;
end
endmodule
Generally, in Verilog/SystemVerilog you cannot assign to a variable from two parallel blocks (with some exceptions). You are assigning to R and Q from two places: the initial block and the always_comb block.
Although the initial block only runs once, it runs in parallel with the always_comb block at the beginning of the simulation, which is a violation of this rule.
Why don't you get rid of the initial block and do everything in always_comb?
always_comb
begin
Q = 8'd0; // set initial value of Q
R = y; // set initial value of R
.... //THE REST OF THE ALGORITHM
end
Also, you are missing using a loop!
An important distinction between writing System Verilog (or any HDL) and writing in any software language (C/C++, Java, etc) is that System Verilog is designed to facilitate describing hardware structures while allowing for software-like testbenches, while software languages are designed to allow users to give instructions to an interpreter, VM or actual hardware. That being said, you need to think first about the hardware you are trying to describe and then write the associated HDL code. There are numerous posts describing the differences between HDLs and software languages (ex: Can Verilog/Systemverilog/VHDL be considered actor oriented programming languages?).
Looking at your code and flow chart you were given, it appears you are trying to use System Verilog as a programming language rather than a HDL. For example, initial blocks are generally only used in test benches and not in modules themselves. Also, as your algorithm is sequential, it is likely you will need a clock signal and registers, but the way your code lacks both.
Ideally, you should have a good idea of how to design digital hardware systems before trying to code any HDL, as this is the mentality you should have using HDLs. The goal is generally to translate a hardware design into HDL, so understanding digital design will help clarify alot.
Related
I understand that always block can be used to implement procedural and sequential logic.
Will the gate-level realization of the following two codes be the same? If yes, what is the correct way of describing this continuous-time logic?
a.
module func(input a, input b , output reg o);
always #(a,b)
o=a&b;
endmodule
b.
module func(input, a, input b, output o);
assign o = a & b;
endmodule
In (a), 'o' is a reg type and in (b) it is a wire. What does this difference mean?
What are the required 'always' block properties for the synthesis tool to implement a FF? I know the following will result in a FF:
always #(posedge clk or negedge rst)
[...]
But, I'm looking for a more in-depth understanding.
Your (a) and (b) codes are functionally equivalent. This means that they will simulate the same way and they will infer the same logic when synthesized.
They use 2 different styles of modeling: (a) uses a procedural assignment because it uses an always block, whereas (b) uses a continuous assignment because there is no always block and it uses the assign keyword.
In this simple case, there is no "correct" way; they are 2 different styles to achieve the same functionality. For such simple logic, it is preferable to use (b) because it uses less code and it is easier to understand. However, if your combinational logic were more complicated, the procedural approach might be easier to understand.
In (a), the o signal must be a reg type since it is assigned inside a procedural logic block. There is no such requirement for continuous assignments. In this case, defining a reg type does not result in a flip-flop. A reg only infers a flip-flop when the always block describes sequential logic
Synthesis tools look for specific types of patterns in Verilog code to infer sequential logic. The following will infer flip-flops with an asynchronous reset:
always #(posedge clk or negedge rst)
if (!rst) ...
else ...
The following will infer flip-flops with an synchronous reset:
always #(posedge clk)
if (!rst) ...
else ...
These are just a couple examples.
I understand that always block can be used to implement procedural and sequential logic.
Will the gate-level realization of the following two codes be the same? If yes, what is the correct way of describing this continuous-time logic?
a.
module func(input a, input b , output reg o);
always #(a,b)
o=a&b;
endmodule
b.
module func(input, a, input b, output o);
assign o = a & b;
endmodule
In (a), 'o' is a reg type and in (b) it is a wire. What does this difference mean?
What are the required 'always' block properties for the synthesis tool to implement a FF? I know the following will result in a FF:
always #(posedge clk or negedge rst)
[...]
But, I'm looking for a more in-depth understanding.
Your (a) and (b) codes are functionally equivalent. This means that they will simulate the same way and they will infer the same logic when synthesized.
They use 2 different styles of modeling: (a) uses a procedural assignment because it uses an always block, whereas (b) uses a continuous assignment because there is no always block and it uses the assign keyword.
In this simple case, there is no "correct" way; they are 2 different styles to achieve the same functionality. For such simple logic, it is preferable to use (b) because it uses less code and it is easier to understand. However, if your combinational logic were more complicated, the procedural approach might be easier to understand.
In (a), the o signal must be a reg type since it is assigned inside a procedural logic block. There is no such requirement for continuous assignments. In this case, defining a reg type does not result in a flip-flop. A reg only infers a flip-flop when the always block describes sequential logic
Synthesis tools look for specific types of patterns in Verilog code to infer sequential logic. The following will infer flip-flops with an asynchronous reset:
always #(posedge clk or negedge rst)
if (!rst) ...
else ...
The following will infer flip-flops with an synchronous reset:
always #(posedge clk)
if (!rst) ...
else ...
These are just a couple examples.
I am trying to code and synthesize in Verilog. Sometimes I am still getting confused with using Verilog as a typical C like programming language, I am trying to understand if there will be a difference between 2 different codes:
always # (a1,a0,b1,b0)
begin
case ({a1,a0,b1,b0})
4'b0000 : s= 7'b1110111 ;
4'b0001 : s= 7'b1110111 ;
....
....
4'b1110 : s= 7'b0101111 ;
4'b1111 : s= 7'b1111010 ;
endcase
end
and
using the above code logic with assign, instead of always block.
Will the above code generate a latch? In which case would it generate a latch? Will there be any delay difference between using the 2 codes?
PS we are trying to create a 2bit binary multiplier, which outputs to a 7 segm display.
Thank you.
Latches are generated when one or more paths through a conditional statement are unassigned. For example:
reg [1:0] a;
reg b;
always#(*)
case (a)
0: b=0;
1: b=0;
2: b=1;
endcase
would generate a latch, because I did not cover the a=3 case. You can avoid this by either explicitly covering each case (like it appears you have done) or by using the default case.
For assign statements, it depends on how you format them, but you're significantly less likely to accidentally infer a latch. For example, if you you the ternary operation (i.e. assign b = a? 1:0;) both halves of the if are inferred.
As for delay, case vs. assign should create the same netlist, so they should produce similar or identical results, provided they're logically identical.
(As a side note, it's good practice to use always#(*), rather than always # (a1,a0,b1,b0); The synthesis tool can figure out the correct sensitivity list.)
I am reading the IEEE Standard Verilog Hardware Description Language (specifically IEEE Std 1364-2001) which unambiguously defines and discusses simulatable Verilog. Unfortunately, the document does not touch upon the notion of synthesis.
I haven't been able to find a similar reference for synthesisable Verilog. All I find is vague rules, or unnecessarily restrictive ones.
Where can I learn the formal language of synthesisable Verilog?
IEEE 1364.1 is an adjunct to the 1364 Verilog standard titled Verilog Register Transfer Level Synthesis, which attempts to define a common synthesizable subset. However, as Jerry points out, different tools support different constructs, and to determine tool-specific behavior, you need to consult the tool documentation.
There isn't a formal (BNF-style) syntax definition for synthesizable Verilog. Whether code is synthesizable depends on usage as well as syntax. For example, the behavior described by an always construct with incomplete sensitivity, like always #(a) o = a || b, isn't synthesizable. (Most tools will synthesize that code as if the sensitivity list were complete, resulting in a possible simulation/synthesis mismatch.)
Circuit constructs like latches and multiply-driven nets can be synthesized from a Verilog description, but are disallowed or discouraged under most design rules. There are also synthesizable constructs that are unsupported or inadvisable given the choice of target library. For example, describing a RAM that's larger than the maximum supported by a chosen FPGA technology, or describing tri-state drivers when they aren't present in the target library.
The general constructs to stick to for synthesizable Verilog are:
Combinational logic modeled with continuous assignments (assign statements)
Combinational logic modeled with always blocks, which should use blocking assignments, and either have a complete sensitivity list or use always #*
Sequential logic (flip-flops) modeled with always blocks, which should use non-blocking assignments, and have either posedge clock alone (for sync reset or non-reset flops) or posedge clock and posedge or negedge reset (for async reset flops) in the sensitivity list.
The safest coding style for sequential logic is to code only reset logic in the sequential always block:
always #(posedge clk or posedge reset)
if (reset)
q <= reset_value;
else
q <= next_value;
However, if you're careful, you can code additional combinational logic in the sequential block. A common case where it may make sense to do this is a mux in front of a flop:
always #(posedge clk)
if (!sel)
q <= sel0_value;
else if (sel)
q <= sel1_value;
else
q <= 'bx;
I'm a bit confused about what is considered an input when you use the wildcard #* in an always block sensitivity list. For instance, in the following example which signals are interpreted as inputs that cause the always block to be reevaluated? From what I understand clk and reset aren't included because they dont appear on the right hand side of any procedural statement in the always block. a and b are included because they both appear on the right hand side of procedural statements in the always block. But where I'm really confused about is en and mux. Because they are used as test conditions in the if and case statements are they considered inputs? Is the always block reevaluated each time en and mux change value? I'm pretty much a noob, and in the 3 Verilog books I have I haven't found a satisfactory explanation. I've always found the explanations here to be really helpful. Thanks
module example
(
input wire clk, reset, en, a, b,
input wire [1:0] mux,
output reg x,y, z
);
always #*
begin
x = a & b;
if (en)
y= a | b;
case(mux)
2'b00: z = 0;
2'b01: z = 1;
2'b10: z = 1;
2'b11: z = 0;
endcase
end
endmodule
Any signal that is read inside a block, and so may cause the result of a block to change if it's value changes, will be included by #*. Any change on a read signal used must cause the block to be re-evaluated, as it could cause the outputs of the block to change. As I'm sure you know, if you hadn't used #* you'd be listing those signals out by hand.
In the case of the code you've provided it's any signal that is:
Evaluated on the right hand side of an assignment (a and b)
Evaluated as part of a conditional (en and mux)
...but it's any signal that would be evaluated for any reason. (I can't think of any other reasons right now, but maybe someone else can)
clk and reset aren't on the sensitivity list because they aren't used. Simple as that. There's nothing special about them; they're signals like any other.
In your example, the following signals are included in the implicit sensitivity list:
a
b
en
mux
clk and reset are not part of the sensitivity list.
This is described completely in the IEEE Std for Verilog (1800-2009, for example). The IEEE spec is the best source of detailed information on Verilog. The documentation for your simulator may also describe how #* works.
The simplest answer depends on if you are writing RTL, or a testbench. If you are writing RTL then you should try to forget about the concept of Sensitivity lists, as they don't really exist. There is no logic that only updates when an item on the list is triggered. All sensitivity lists can do in RTL is cause your simulation and actual circuit to differ, they don't do anything good.
So, always use "always #*" or better yet "always_comb" and forget about the concept of sensitivity lists. If the item in the code is evaluated it will trigger the process. Simple as that. It an item is in an if/else, a case, assigned to a variable, or anything else, it will be "evaluated" and thus cause the process to be triggered.
But, just remember, in digital circuits, there is no sensitivity list.