verilog code for square of a number without using * operator

[alangs]

can anybody tell me how i can calculate the square of a number without using * operator.plz give me the answer in verilog code by using for loop......

 

[bigdogguru]

Re: verilog code for square of a number without using * oper

Algorithm is simple:

2^2 = 2+2 = 4

3^2 = 3+3+3 = 9

4^2 = 4+4+4+4 = 16

5^2 = 5+5+5+5+5 = 25

C code:

int square(int x)
{
      int sum = 0;

      for(int count = x; count>0; count--)
      {
         sum = sum + x;
      }

      return sum;

}

Translate into your preferred HDL.

 

[sameh_yassin99]

Re: verilog code for square of a number without using * oper

Hello,

It is better to use a multiplierso that you use combinational logic. However, if you like to use pipelining in your "square" circuit, you should write code for something like the diagram below. Note that this is a vaiable delay circuit, which means that you'll wait for 10^2 longer than 5^2



Best Wishes,