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sys. verilog code - need explanation

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HI, I am new to s.verilog - any help with the code below will be great!
what does the code below actually mean? anyone explain plz?

why is LOC_IDLE triggered?

LOC :
        begin
          varc_sel[3:0] = { LOC_DRAIN ,
                                  LOC_GETDEL ,
                                  LOC_BUFDEL ,
                                  LOC_IDLE } ;
          case (varc_sel[3:0])
            4'b1000 ,
            4'b0100 ,
            4'b0010 ,
            4'b0001 ,
            4'b0000 :
              begin
               
              end
            default :
              begin
                 $write("ERROR:");
                 $display(" Invalid arc in, fsm:SM, state:LOC") ;
                 $display(" Instance: %m");
                 $display(" Time: %t", $time);


Originally posted in cdnusers.org by indeb
Parents

  • Yes, you are correct. The way the case works is it looks at the value of varc_selA and then selects the branch that matches the value. In the example, if the value is 8, 4, 2, 1, or 0 it will branch to the 'begin...end'. If it doesn't match any branch, then the default branch will be executed.

    I'm reposting my previous text since the formatting got screwed up.

    Somewhere I assume varc_sel is defined to be an enum like:

    typedef enum bit [2:0] {LOC_DRAIN, LOC_GETDEL, LOC_BUFDEL, LOC_IDLE} loc_t;
    loc_t varc_sel;

    The way enums work is that you can refer to the labels by name, but underneath there are actual values for those labels. By default, it starts with 0 and counts up. You can change the default behavior by providing your own values in the definition. If you used the defaults, then LOC_IDLE would be equal to 3 which would never be selected in the case. However, you could use the following enum definition to make the assignment one-hot:

    typedef enum bit [2:0] {LOC_DRAIN=1, LOC_GETDEL=2, LOC_BUFDEL=4, LOC_IDLE=8} loc_t;


    Originally posted in cdnusers.org by tpylant
Reply

  • Yes, you are correct. The way the case works is it looks at the value of varc_selA and then selects the branch that matches the value. In the example, if the value is 8, 4, 2, 1, or 0 it will branch to the 'begin...end'. If it doesn't match any branch, then the default branch will be executed.

    I'm reposting my previous text since the formatting got screwed up.

    Somewhere I assume varc_sel is defined to be an enum like:

    typedef enum bit [2:0] {LOC_DRAIN, LOC_GETDEL, LOC_BUFDEL, LOC_IDLE} loc_t;
    loc_t varc_sel;

    The way enums work is that you can refer to the labels by name, but underneath there are actual values for those labels. By default, it starts with 0 and counts up. You can change the default behavior by providing your own values in the definition. If you used the defaults, then LOC_IDLE would be equal to 3 which would never be selected in the case. However, you could use the following enum definition to make the assignment one-hot:

    typedef enum bit [2:0] {LOC_DRAIN=1, LOC_GETDEL=2, LOC_BUFDEL=4, LOC_IDLE=8} loc_t;


    Originally posted in cdnusers.org by tpylant
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