June 7th 2006

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The 4 stages in floating point addition pipeline are:
1.Exponent comparer
2.Mantissa shifter
3.Mantissa adder
4.Normalize result

// Aim:To Implement 4 stage floating point addition pipeline import java.util.*; class fp // class of floating pt. nos. { double man; int exp; } class fp_add { static int compare_exp(fp f1,fp f2) // compares the exponent { if(f1.exp<f2.exp diff="f2.exp-f1.exp;" claculates difference in exponent return represents first number else if second both exponents are same static fp shift_man f shifts the mantissa by using value for i="0;i<diff;++i)" f.man f.exp normalize final result while>=10) { f.man/=10; f.exp+=1; // exponent is -ve and hence adding 1 will increase its val ex. -5+1=-4 } return f; }  static int diff; // difference in exponent values public static void main(String args[]) { Scanner scr=new Scanner(System.in);  /* creating objects for two floating pt. nos. and result */ fp f1=new fp(); fp f2=new fp(); fp r=new fp();  System.out.println("Enter mantissa and exponent of x "); f1.man=scr.nextDouble(); f1.exp=scr.nextInt(); System.out.println("Enter mantissa and exponent of y "); f2.man=scr.nextDouble(); f2.exp=scr.nextInt();  int choice=compare_exp(f1,f2); System.out.print("STAGE 1(exponent comparer): The num with smaller exponent is "); if(choice==1)System.out.println("x  \n");else if(choice==2)System.out.println("y  \n");else System.out.println("Both are same.\n");  System.out.print("STAGE 2(mantissa shifter): ");  if(choice==1) // choice is 1 hence first mantissa has to be shifted { f1=shift_man(f1); System.out.println("shifted mantissa "+f1.man+"\n"); } else if(choice==2) // choice is 2 hence second mantissa has to be shifted { f2=shift_man(f2); System.out.println("shifted mantissa "+f2.man+"\n"); } else // equal exponents System.out.println("no shifting required\n");  System.out.print("STAGE 3(mantissa adder): "); r.man=f1.man+f2.man; System.out.println("added mantissa "+r.man+"\n"); r.exp=f1.exp; // both f1 and f2 NOW have same exponents.Hence assign any one's exponent.    r=normalize(r); // call the method to normalize final result. System.out.println("STAGE 4(normalize result):"); System.out.println("The normalized result is "+r.man+" * 10^"+r.exp);  int c=-1*r.exp;  /* Additional code to represent answer in different form ex 3.52*10^-3 will be 0.00352 (int i=0;i<c r.man="r.man/10;" system.out.println the result is c:>java fp_add Enter mantissa and exponent of x 8.5 -3 Enter mantissa and exponent of y 2.345 -5 STAGE 1(exponent comparer): The num with smaller exponent is y  STAGE 2(mantissa shifter): shifted mantissa 0.023450000000000002  STAGE 3(mantissa adder) added mantissa 8.52345  STAGE 4(normalize result): The normalized result is 8.52345 * 10^-3 */ </c></f2.exp>

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