The below program calculates 2⁶² + 2⁶¹ + 2⁶⁰ + … + 2¹ + 2⁰.

Using double type to store sum:

double sum = 0;
  for (int i = 0; i < 63; i++) {
    sum += pow(2.0, i);
    // print("i : $i sum : $sum");
  }
  print(sum);

OUTPUT:

9223372036854776000.0

It is clear that answer should be an odd number but the sum is : 9223372036854776000.0 which is an even number.

When I change the sum to int data type it’s giving the accurate result:

Using int type to store sum:

int sum = 0;
for (int i = 0; i < 63; i++) {
  sum += pow(2.0, i).toInt();
  //print("i : $i sum : $sum");
}
print(sum);

OUTPUT:

9223372036854775807

When the datatype is double why the summation is generating wrong result ?

>Solution :

Because a double only has 52 significant bits in the mantissa, whereas your number has 63.