Question regarding coverage

SystemVerilog
1

Hi There,

lets say I have 3 bit counter I need to write a coverage does it cover all possible values. with less bins for example we can write 8 single bins to cover. but what if if counter is 32 bit to cover all possible values with less bins.

I answer this question with various syntax in coverage but still it is not the answer.
for example

bins [] = [0:8] // it will give nine separate bins
transition bins for 3 bit I can write like this. what if I have 32 bits you cannot able to write all possible combination.

2

Put a number inside the the bins []

  bit [31:0] counter;
  covergroup cg;
    coverpoint counter {
      bins counter_range[64] = {[0:$]};
    }
  endgroup