How to declare and use 1D and 2D byte arrays in Verilog? - Stack ...

Verilog thinks in bits, so reg [7:0] a[0:3] will give you a 4x8 bit array (=4x1 byte ... i++) begin a[i] = i[7:0]; end c = a[0]; d = a[1][2]; // using 2D for (int i=0; i

stackoverflow.com