$size, $bits, verilog

问题:

What is the difference between $size and $bits operator in verilog.? if I've variables, [9:0]a,[6:0]b,[31:0]c.

c <= [($size(a)+$size(b)-1]-:$bits(b)];

What will be the output at 'c' from the above expression?

回答1:

$size shall return the number of elements in the dimension, which is equivalent to $high - $low + 1. It is relative to the dimension, not only bit counts. If the type is 1D packed array or integral type, it is equal to $bits.

$bits system function returns the number of bits required to hold an expression as a bit stream.

$bits ( [expression|type_identifier] )

It returns 0 when called with a dynamically sized type that is currently empty. It is an error to use the $bits system function directly with a dynamically sized type identifier.

I have no idea about your question, c <= [($size(a)+$size(b)-1]-:$bits(b)];. Is it a valid expression in RHS? Are you talking about the array range expression, [n +: m] or [n -: m] ?

回答2:

$size() gives the number of bits for a single dimension. $bits() gives the number of bits to completely represent the variable.

For example:

reg [9:0] a;
reg [9:0] b [5:0];

initial begin
  $display("a Size ", $size(a));
  $display("a Bits ", $bits(a));
  $display("b Size ", $size(b));
  $display("b Bits ", $bits(b)) ;
end

Gives :

a Size          10
a Bits          10
b Size           6 // Depth of memory
b Bits          60 // Width * Depth

In your case you just have 1 dimensional arrays, not memories or structs so $size() and $bits() would be the same thing.