Verilog: differences between if statement and case statement

Question

I am new to Verilog language and want to do some practices to get familiar with it. And I encountered this problem on HDLbits: DFF8ar

This problem asks me to create 8 D flip-flops with active-high asynchronous reset. I use a case statement to handle the areset signal:

module top_module (
input clk,
input areset,   // active high asynchronous reset
input [7:0] d,
output reg[7:0] q
);

always @(posedge clk or posedge areset) begin
    case (areset)
        1'b1: q <= 8'b0; 
        default: q <= d;
    endcase
end
endmodule

And to my surprise, the circuit it generated ignores the clk signal:

But, if I switch the case statement to an if-else statement, the result will be correct:

always @(posedge clk or posedge areset) begin
    if (areset)
        q <= 8'b0; 
    else q <= d;
end

I don't know the reason behind it even after doing some research. Does if-else statements and case statements have some fundamental difference? Any help is appreciated!

toolic · Accepted Answer · 2021-07-11T10:03:20.233

Synthesis places some special restrictions on the normal Verilog language. Synthesis tools recognize specific Verilog coding patterns, but your case code does not match any of those patterns, whereas your if/else code does. There should be documentation with your tool set that shows what syntax is supported for synthesis. When you ran the synthesis, there may have been warning or error messages; check for any log files.

Although the 2 coding styles may behave the same in simulation, you need to restrict yourself to supported syntax for synthesis.

Verilog: differences between if statement and case statement

1 Answers1