Is the ++ operator in System Verilog blocking or non-blocking?

Question

Good coding convention says that we should use blocking assignments in a combinational block, and non-blocking assignments in a sequential block. I want to use the ++ operator in a combinatorial block, but I don't know if it is blocking. So is this code:

input [3:0] some_bus;
logic [2:0] count_ones;
always_comb begin
  count_ones = '0;
  for(int i=0; i<4; i++) begin
    if(some_bus[i])
      count_ones++;
  end
end

equivalent to this:

input [3:0] some_bus;
logic [2:0] count_ones;
always_comb begin
  count_ones = '0;
  for(int i=0; i<4; i++) begin
    if(some_bus[i])
      count_ones = count_ones + 1;
  end
end

or this:

input [3:0] some_bus;
logic [2:0] count_ones;
always_comb begin
  count_ones = '0;
  for(int i=0; i<4; i++) begin
    if(some_bus[i])
      count_ones <= count_ones + 1;
  end
end

I did look in the 1800-2012 standard but could not figure it out. An answer that points me to the appropriate section in the standard would be appreciated.

dwikle dwikle · Accepted Answer · 2014-03-19T14:54:09

According to section 11.4.2 of IEEE Std 1800-2012, it is blocking.

SystemVerilog includes the C increment and decrement assignment operators ++i , --i , i++ , and i-- . These do not need parentheses when used in expressions. These increment and decrement assignment operators behave as blocking assignments.

Is the ++ operator in System Verilog blocking or non-blocking?

1 Answers