I am new to Verilog, I was trying to write a simple code but I am not sure how to do it in a expert way. I have a 12 bit register "data", each bit of that register have a specific value. e.g.
Bit 0 = 12;
Bit 1 = 16;
Bit 2 = 33;
......
Bit 11 = 180;
Now if any bit of "data" register is 1 then the result should be the sum of all value that coresponds to that bit value. e.g.
data = 12'b100000000101
result = 225 (180+33+12)
Right now i am checking each bit of data, if it is 1 then i register that corresponding value and add it to previous registered value. This method takes number of cycles. How can i do it in a fast way in verilog.
thank you
It depends on what you mean by "fast". Presumably you mean time, but remember that time=cycles/frequency - reducing the number of cycles will often reduce the maximum frequency your circuit can operate at.
For example, here's a circuit that does the entire add in one cycle:
always@(*) begin
tempsum = 0;
tempsum = tempsum + (data[0]? 12:0);
tempsum = tempsum + (data[1]? 16:0);
tempsum = tempsum + (data[2]? 33:0);
//...
end
always@(posedge clock)
result <= tempsum;
If you synthesized this circuit, you'd see a long chain of adders. In could calculate the result in a single cycle, but would have a long critical path, and therefore have a lower fMax. Whether this would be "faster" is impossible to know until you synthesize it (there are too many factors to guess).
A better multi-cycle approach could be to use a tree, i.e.:
reg [31:0] sum [29:0];
always @ (posedge clock) begin
// level 0
sum[0] <= (data[0]? 12:0) + (data[1]? 16:0);
sum[1] <= (data[2]? 33:0) + (data[3]? 40:0);
// ...
sum[15] <= (data[30]? 160:0) + (data[31]? 180:0);
// level 1
sum[16] <= sum [0] + sum [1];
sum[17] <= sum [2] + sum [3];
// ...
sum[23] <= sum [14] + sum [15];
// level 2
sum[24] <= sum [16] + sum [17];
sum[25] <= sum [18] + sum [19];
// ...
// level 3
sum[28] <= sum [24] + sum [25];
sum[29] <= sum [26] + sum [27];
result <= sum [28] + sum [29];
end
All that said, ultimately the "fastest" approach will also depend on the other requirements of your system, what you're implementing it on, etc.