3 Responses to assinment1..problem in Q-15

1. rahul.gautam says:

   June 26, 2012 at 8:16 pm

   it will be showing some random value….because float data type have precision till e7….after this it starts giving some random value…and in this above equation you are trying to modify 21st digit which is not possible.

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2. arjumand says:

   June 26, 2012 at 9:59 pm

   e20 is the scientific notation……it means 10 to the power 20

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3. rahul gaur says:

   June 26, 2012 at 10:15 pm

   Hi here is my code take a look ..
   I am also skeptic about the answer…

   e-notation basically 2.0e20(or AeN) represents 2.0 * 10^20..

   #include
   #include

   int main(void)
   {
   float a,b;
   b = (2.0*pow(10,20)) + 1.0;//2.0e20 Can be expressed as 2.0 * 10^20
   printf(“Value of b is %f\n”,b);

   float c = 2.0e20;
   printf(“Value of c is %f\n”,c);

   a = b – (2.0*pow(10,20))/*c*/;//if I replace the exp with var c result changes to 0.0000….
   printf(“Value of %f – %f = %f \n”,b,c,a);//Result seems lil Strange

   return 0;

   }

   =============output=================
   for the code in question
   [root@linuxstation asgmt]# ./fourteen
   ./fourteen
   Value of b is 200000004008175468544.000000
   Value of c is 200000004008175468544.000000
   Value of 200000004008175468544.000000 – 200000004008175468544.000000 = 4008175468544.000000

   +++++++++++++output+++++++++++++++++++
   when I replace the a = b – c ;

   [root@linuxstation asgmt]# make fourteen
   make fourteen
   cc fourteen.c -o fourteen
   [root@linuxstation asgmt]# ./fourteen
   ./fourteen
   Value of b is 200000004008175468544.000000
   Value of c is 200000004008175468544.000000
   Value of 200000004008175468544.000000 – 200000004008175468544.000000 = 0.000000
   [root@linuxstation asgmt]#

   Reply