For e.g.
initial
begin
generate
for(genvar i; i < 4; i++)
//Code
endgenerate
end //initial
I'm getting error using QuestaSim with the concept. "Near generate: syntax error, unexpected generate "
No. generate blocks are evaluated during elaboration time. While initial,always and other procedural blocks start at zero simulation time, that is, run-time. Referring to Systemverilog IEEE 1800-2012 :
Generate schemes are evaluated during elaboration of the design.
Although generate schemes use syntax that is similar to behavioral
statements, it is important to recognize that they do not execute at
simulation time.
They are evaluated at elaboration time, and the
result is determined before simulation begins. Therefore, all
expressions in generate schemes shall be constant expressions,
deterministic at elaboration time.
In Verilog, instantiating a module means adding extra hardware to the board.
This hardware must be added before simulation starts(i.e. at compile time). You can not add/remove hardware during run time. You can either conditionally instantiate a module or multiply instantiate it, but never at run time.
Refer generate block syntax error question for an idea about your error. Also, refer this question for generate and genvar understanding. Referring IEEE 1800-2012 Chapter 27 for more information.
EDIT:
To create and pass multiple interface instances, the total number of interface instances must be governed by some parameter or macro. You can use this parameter in for loop in generate block to create distinct instances and set each of them using different key as follows:
// Generate multiple instances of interface
genvar i;
generate
for(i=0;i<NUM_OF_INTERFACES;i++)
begin
// Generate clk with different period for each instance
always #(i+1) clk[i] = ~clk[i];
inter in(clk[i]); // Create multiple instances here
initial
begin
// Set each and every instance
uvm_config_db#(virtual inter)::set(null,"*",$sformatf("in_%0d",i),in);
end
end
endgenerate
A complete example is created at EDAPlayground Multiple Interface link. Creating multiple instances can be referred from this question.
Related
This is a system verilog file .The below code when run gives an error "Illegal operand for constant expression" at "if(m_val !==0) and if(div_val)"
Have tried without generating statements but still the same error. Any help on what is wrong with the code?
logic m_ref_clk,mul_ref_clk,div_ref_clk,mixed_ref_clk;
real clk_period = ((1.0/(freq*BASE_GHz_CLK_FREQ))*1s);
generate
if(m_val !== 0) begin
if(div_val == 0)begin
initial begin
clk_period = clk_period*m_val;
forever begin
#(clk_period) mul_ref_clk = ~mul_ref_clk;
end
end
end
end
endgenerate
assign multiply_clk = mul_ref_clk;
Generate statements are evaluated during elaboration which occurs before t = 0 in a simulation. Therefore normal variables (updated dynamically as the RTL runs) can't be used a arguments to generate statements.
Elaboration from 1800 2012 section 3:
Elaboration occurs after parsing the source code and before simulation;
and it involves expanding instantiations, computing parameter values, resolving hierarchical names,
establishing net connectivity and in general preparing the design for simulation.
Generate statements from IEEE 1800 2012 section 27:
Generate schemes are evaluated during elaboration of the design. Although generate schemes use syntax that
is similar to behavioral statements, it is important to recognize that they do not execute at simulation time.
They are evaluated at elaboration time, and the result is determined before simulation begins. Therefore, all
expressions in generate schemes shall be constant expressions, deterministic at elaboration time. For more
details on elaboration, see 3.12
Say I have a Verilog module that's parameterizable like the below example:
// Crunches numbers using lots of parallel cores
module number_cruncher
#(parameter NUMBER_OF_PARALLEL_CORES = 4)
(input clock, ..., input [31:0] data, ... etc);
// Math happens here
endmodule
Using Verilog 1364-2005, I want to write a testbench that runs tests on this module with many different values NUMBER_OF_PARALLEL_CORES.
One option that I know will work is to use a generate block to create a bunch of different number_crunchers with different values for NUMBER_OF_PARALLEL_CORES. This isn't very flexible, though - the values need to be chosen at compile time.
Of course, I could also explicitly instantiate a lot of different modules, but that is time consuming and won't work for the sort of "fuzz" testing I want to do.
My questions:
Is there a way to do this by using a plusarg passed in from the command line using $value$plusargs? (I strongly suspect the answer is 'no' for Verilog 1364-2005).
Is there another way to "fuzz" module parameterizations in a testbench, or is using a generate block the only way?
Since $value$plusargs is evaluated at runtime, it can not be used to set parameter values, which must be done at compile-time.
However, if you use generate to instantiate multiple instances of the design with different parameter settings, you might be able to use $value$plusargs to selectively activate or enable one instance at a time. For example, in the testbench, you could use the runtime argument to only drive the inputs of a specific instance.
I have written the below code for a simple multiplication of 2 n-bit numbers(here n=16). It is getting simulated with desired output waveform but the problem is, it is not getting synthesized in vivado 17.2 even though I have written a static 'for loop'(ie., loop iteration is constant). I am getting below mentioned error.
[Synth 8-3380] loop condition does not converge after 2000 iterations
Note: I have written
for(i=i;i<n;i=i+1)
instead of
for(i=0;i<n;i=i+1)
because the latter one was executing once again after i reached n. So that is not a mistake. Kindly someone help. Thank you for your time
//unsigned integer multiplier
module multiplication(product,multiplier,multiplicand,clk,rset);
parameter n = 16;
output reg [(n<<1)-1:0]product;
input [n-1:0]multiplier, multiplicand;
input clk,rset;
reg [n:0]i='d0;
always #( posedge clk or posedge rset)
begin
if (rset) product <= 'd0;
else
begin
for(i=i;i<n;i=i+1)
begin
product =(multiplier[i] == 1'b1)? product + ( multiplicand << i ): product;
$display("product =%d,i=%d",product,i);
end
end
end
endmodule
First of all, it is not a good practice to use for, while kind of loops if you really want to implement your design on FPGA (Vivado is optimized to be used to implement your design on FPGA). Even if you can successfully synthesize your design, you might face with timing problems or unexpected bugs.
I think you can find your answer here.
Edit: I just wanted to inform you that generally controlling timing is very important in HW design, especially when you want to integrate your design with other system, loops can be nightmare for that.
Go back to using your original for (i=0 loop.
Your error is that you assume i=0 because of reg [n:0]i='d0; That is only true the very first time. Thus only once, at the start of the simulation.
because the latter one was executing once again after i reached n.
Yes, the loop will repeat again and again for every clock cycle. That is what #( posedge clk ...) does.
More errors:
You are using blocking assignment in the clock section, use non-blocking:
product <=(multiplier[i] == 1'b1)? product + ( multiplicand << i ): product;
Your product is only correct the first time after a reset (when it starts at zero). The second clock cycle after a reset you do the multiplication again but start with the previous value of product.
Your i is a bit big you use 17 bits to count to 16. Also global loop variables have pitfalls. I suggest you use the system Verilog syntax of: `for (int i=0; ....)
The error message is not about the iterations of your loop, but about the iterations of the internal algorithm of your synthesis program. It tells you that Vivado failed to create a circuit that could actually be implemented on your FPGA of choice at your clock speed of choice and which actually does what you're asking for.
Let me elaborate, after I mention two items of general import: don't use blocking assignments (=) inside always #(posedge clk) blocks. They almost never do what you want. Only non-blocking assignments (<=) should be clocked. Secondly, the correct way to synthesize a for loop is with generate, even though Vivado seems to accept the simple for.
You are building a fairly large block of combinatorial logic here. Remember that when you synthesize combinatorial logic you are asking for a circuit that can evaluate the expression that you've written within a single clock cycle. The actual expression taken into consideration is the one after the for loop has been unrolled. I.e., you are (conditionally) adding a 16bit number to a 32bit number 16 times, each times shifted one bit further to the left. Each of these additions has a carry bit. So each of these additions will actually have to look at all of the upper 16bits of the result of the previous addition, and each will depend on all but the lowest bit of the previous addition. An adder for one bit + carry needs O(5) gates. Each addition is conditional, which adds at least one more gate for each bit. So you are asking for at minimum of 16*16*6 = 1300 interdependent gates that all have to stabilize within a single clock cycle. Vivado is telling you that it cannot fulfill these requirements.
One way of mitigating this problem will be to synthesize for a lower clock frequency, where the gates have more time to stabilize, and you can thus build longer logic chains. Another option would be to pipeline the operation, say by only evaluating what corresponds to four iterations of your loop within a single clock cycle and then building the result over several clock cycles. This will add some bookkeeping logic to your code, but it is inevitable if one wants to evaluate complex expressions with finite resources at high clock frequencies. It would also introduce you to synchronous logic, which you will have to learn anyway if you want to do anything non-trivial with an FPGA. Note that this kind of pipelining wouldn't affect throughput significantly, because your FPGA would then be doing several multiplications in parallel.
You may also be able to rewrite the expression to handle the carry bits and interdependencies in a smarter way that allows Vivado to find its way through the expression (there probably is such a way, I assume it synthesizes if you simply write the multiplication operator?).
Finally, many FPGAs come with dedicated multiplier units because multiplication is a common operation, but implementing it in logic gates wastes a lot of resources. As you found out.
I have a custom module in Verilog that takes several inputs and a few parameters.
module modA (
input inp1, input inp2, ... output out);
parameter my_addr;
...
endmodule
For various reasons associated with the specific hardware that I'm using, I need to create many instances of this module, each time with a different parameter (because my_addr corresponds to a specific location in memory that I am then accessing from the host). Moreover, I would like an easy way to access all of the outputs. I thought of doing something like
wire [9:0] outputs;
modA #(.my_addr) all_modA[9:0](.inp1(inp1), .inp2(inp2), .out(outputs));
So then I would have a vector of elements of all_modA, and the nth element of outputs would correspond to the nth instantiation of modA. That sounds like it would work well, except I need to pass a unique address to each instantiation. When I was just doing one instantiation at a time, it was quite easy to just define a parameter, but I don't see how I could pass a unique identifier each time.
Is there any way to accomplish this? The problem is, if I use a for loop, then I don't see how I can instantiate each modA with a unique identifier. If I don't use a for loop, then I'm not sure how to add a unique identifier.
You can use a generate block over here. Generate blocks are evaluated during elaboration of the design. They do not execute at simulation time and the result is determined before the simulation begins. Therefore, all expressions in generate schemes shall be constant expressions, deterministic at elaboration time.
Give different values to my_addr according to the requirement. Also, you can apply some label identifier to begin-end block. This as an optional feature for generate blocks, but very highly recommended. By doing so, a unique hierarchical path can be seen to every instance of modA is different for each module.
For your code, I have created a testbench that instantiates the module with changing value of my_addr and outputs bus driving the out signal.
module modA #(parameter int my_addr) (input inp1, input inp2,output out);
initial $display("%m: %d",my_addr);
endmodule
module top();
wire [9:0] outputs;
genvar i; // generate iterator
generate
for(i=0;i<9;i++) begin: some_identifier
// my_addr = i+1 for every module
modA #(.my_addr(i+1)) all_modA(.inp1(inp1), .inp2(inp2), .out(outputs[i])); // Multiple modules
end
endgenerate
endmodule
For more information, refer to IEEE 1800-2012 Section 27. Similar questions can be found here and here.
I am having a bit of trouble instantiating a module in verilog. I am using the Altera Quartus platform to develop and simulate the verilog code.
I have followed this example (among several others):
http://www.asic-world.com/verilog/verilog_one_day4.html
I have written a module (maximum) which finds the maximum between two signed inputs.
Another module I am developing is a systolic array for genetic sequence alignment. The details are not important, however when I try to instantiate a maximum module I get an error.
This is my code so far:
module maximum (a, b, out);
input signed [15:0] a;
input signed [15:0] b;
output reg signed [15:0] out;
always #* begin
if (a>b)
assign out = a;
else
assign out = b;
end
endmodule
and I instantiate in another module systolic_PE (all of this is in the same file seqalign.v)
maximum m0(.a(tempB), .b(diag), .out(tempA));
And I get the error :
'Verilog HDL syntax error at seqalign.v(139) near text "m0"; expecting
"<=" or "="'
I checked everything I have done so far, and I cant seem to see anything I have missed out on.. could anyone be kind enough to guide me?
Also on a side note:
Instantiation of a module in verilog
I was trying to instantiate my module in a if statement, so I tried outside of the if statement in a always #(posedge clk) block, and I get the error
HDL syntax error at seqalign.v(88) near text "("; expecting ";"
Looking over the code you posted in your comment, the issue is from instantiating your module inside your always #(posedge clk) block on line 70. You never instantiate modules inside of procedural blocks (always, initial, etc).
As Verilog is a Hardware Descriptive Language, you have to be in the mindset of designing hardware when writing your code. Module instantiation is like soldering a chip onto a PCB, at design time you either do it, or you dont, and that component stays there for all time. You dont say, well, I want this chip here some of the time, but take it off the PCB when the system gets into these states. In your code, you conditionally instantiate your module if state is 3. However, state changes over time. So that is akin to saying, when the register containing state reads 3, place down this chip into the system, otherwise, it doesnt exist and take it out. On a code level, think of instantiated modules as their own procedural blocks, just as you dont put always inside of other always, dont put modules in always blocks (of course, module definitions/declarations can have always blocks inside them).
Modules are persistent and compile time constant, so you can use generates to conditionally instantiate modules at compile time (ie, decide whether or not to include the module in the design when building the system). But in your code, you are conditionally instantiating at simulation time, which is not allowed as described above.
You can do one of two things to solve your problem. One would be to move your task from your submodule maximum into the systolic_PE module and use it to get the maximum of your variables tby calling it (line 123 would become something like tempA <= convert(.a(0), .b(diag+match)); with a and b added as inputs to your task). Or, instantiate the module outside the always block, but youll need to change your task to be a procedural block like you have in the actual post.