Verilog - Array of Inputs

Question

Im new to Verilog, so please just dont blast me.

If in a module i declare as inputs' vector

[0:3]D

or

[3:0]D

What does it change ?

• The Photon: 1. Verilog or SystemVerilog? 2. Do you want a vector or an array? In Verilog they're two different things.

Answer 1

Both are allowed.

[0:3] D

Is the 'big endian' convention. However 99.99%** of all Verilog code uses the little endian convention when declaring ports and signals even when building a big-endian processor.

The other convention is that the LS bit has index 0. For example if you have a 32-bit wide address bus the convention is to use:

wire [31:0] address_bus;

This does not mean you can't use it. You might want to use only the top 30 bits in a module in which case it is perfectly normal to make an input bus:

input [31:2] address_bus_top;

In fact when I see that in a module which I get from one of my colleagues (Who all follow these conventions) it tells me that it is a sub-set of a bigger vector.

Last if working with memories (2 dimensional arrays) the index convention is:

reg [7:0] byte_mem [0:255];

I have seen warnings from Cadence compilers if that convention is not followed. I think it happens if you use it in combination with $readmem...

**Rough estimate.

Answer 2

It changes the meaning how the MSB and LSB get indexed. D[0] would be the MSB or LSB of the bit vector depending on which declaration you use. If you never reference a bit or part select of a vector, it does not make any difference.