Why doesn't this loop run for i = 1?

Question

This loop is not running for i = 1
It should run 100 times but it's only running 99 times

#include<iostream>
using namespace std;
int main() { 
   double i = 0.01, j = 0.01;
   while(i<=1){
       cout<<i<<' ';
       i = i+j;
   } 
return 0; 
}

don't use floating point numbers for loop conditions. Use an integer and then inside the loop `i = 0.01 * int_index;` — 463035818_is_not_an_ai, Jun 16 '21 at 08:54

score -2 · Answer 1 · answered Jun 16 '21 at 08:54

-2

Why don't start with i= 0.00, because 0.01 is already consuming first number. So 99 left.

answered Jun 16 '21 at 08:54

faheem999

52
3

1

If i = 0.00 the loop should run 101 times. – Harikesh Pratap Singh Jun 16 '21 at 09:02

Why doesn't this loop run for i = 1?

1 Answers1