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Don’t expect @fragnen to do that. Welcome to the world obtuse fragnen posts.These are questions that can be easily answered by trying it, or searching examples. You are not going to get far learning Verilog by asking one question at time.
always @ (posedge clock or negedge rst)
case (rst)
begin
0: q = 1'b0;
1: case (enable)
begin
0: q = q;
1: q = d;
end
end
this style goes against typical coding styles. synthesis tools might complain or might even misinterpret the intent (unlikely). in short, there is no good reason why anyone would ever write this freakshow of a code.
Do you want to mean the following as the template for register interference when you mentioned of register interference?But your construct ignores the template for register inference, as said not a problem of nested case statements.
always@(posedge clock or negedge rst)
if(!rst)
q<=1'b0;
else
q<=d;
Your willful ignorance is simply astounding.Do you want to mean the following as the template for register interference when you mentioned of register interference?
always@(posedge clock or negedge rst)
if(!rst)
q<=1'b0;
else
q<=d;
If the answer of the above question is yes by you, then this above template is one of the fundamental template that is used for coding many types of sequential logic and hence for sequential logic coding in many of the scenarios we cannot use nested case as the usual template for sequential coding is the above template and not the nested case.
it's almost as if you are making an effort to incorrectly understand what we are saying to you over and over. jesus.Do you want to mean the following as the template for register interference when you mentioned of register interference?
always@(posedge clock or negedge rst)
if(!rst)
q<=1'b0;
else
q<=d;
If the answer of the above question is yes by you, then this above template is one of the fundamental template that is used for coding many types of sequential logic and hence for sequential logic coding in many of the scenarios we cannot use nested case as the usual template for sequential coding is the above template and not the nested case.
Code Verilog - [expand] 1 2 3 4 5 6 7 8 9 10 always @ (posedge clock or negedge rst) case (!rst) begin 0: q = 1'b0; 1: case (enable) begin 0: q = q; 1: q = d; end end
I’m done with you, @fragnen.Do you want to mean that if the above code is corrected as below it will be a correct synthesizable code for a D-flipflop with an Enable?
Code Verilog - [expand] 1 2 3 4 5 6 7 8 9 10 always @ (posedge clock or negedge rst) case (!rst) begin 0: q = 1'b0; 1: case (enable) begin 0: q = q; 1: q = d; end end