> what is the error in the following code. in it the main module is
> "test". in that module's "always" block another module "counter" is
> called. but it shows error. how can i solve the problem? how can i
> call another module in always block?
>
Hi there,
First of all, there is a newsgroup for Verilog discussions:
comp.lang.verilog (and I noticed that you just post your
question there as well).
You cannot put a module in a process that only execute
at clock edge. Try think that the counter module is a
hardware, and you can't make it appear only at clock edge,
it exist in the design all the time. So instead of putting it
in a process, you should instantiate it.
module test(clock,reset,count);
input clock, reset;
output [3:0] count;
// reg [3:0] count; // This line is removed because
// it is not controlled by a process.
// Instantiate counter
counter counter1(clock, reset, count);
endmodule
If you want to do a number of processing inside a process,
you could use "function". Google "Verilog" and "function"
should come up with lots of examples.
Hope this help.
regards,
Joseph
what is the error in the following code. in it the main module is
"test". in that module's "always" block another module "counter" is
called. but it shows error. how can i solve the problem? how can i
call another module in always block?
module counter(clock, reset, count);
input clock, reset;
output [3:0] count;
reg [3:0] next_count,count;
always@*
begin
if(count<15)
next_count=count+4'd1;
else
next_count=count;
end
always@(posedge clock)
begin
if(reset)
count<=4'd0;
else
count<=next_count;
end
endmodule
module test(clock,reset,count);
input clock, reset;
output [3:0] count;
reg [3:0] count;
always @(clock)
counter(clock, reset, count);
endmodule