biggest integer that can be stored in a double

The biggest/largest integer that can be stored in a double without losing precision is the same as the largest possible value of a double. That is, DBL_MAX or approximately 1.8 × 10³⁰⁸ (if your double is an IEEE 754 64-bit double). It's an integer. It's represented exactly. What more do you want?

Go on, ask me what the largest integer is, such that it and all smaller integers can be stored in IEEE 64-bit doubles without losing precision. An IEEE 64-bit double has 52 bits of mantissa, so I think it's 2⁵³:

• 2⁵³ + 1 cannot be stored, because the 1 at the start and the 1 at the end have too many zeros in between.
• Anything less than 2⁵³ can be stored, with 52 bits explicitly stored in the mantissa, and then the exponent in effect giving you another one.
• 2⁵³ obviously can be stored, since it's a small power of 2.

Or another way of looking at it: once the bias has been taken off the exponent, and ignoring the sign bit as irrelevant to the question, the value stored by a double is a power of 2, plus a 52-bit integer multiplied by 2^{exponent − 52}. So with exponent 52 you can store all values from 2⁵² through to 2⁵³ − 1. Then with exponent 53, the next number you can store after 2⁵³ is 2⁵³ + 1 × 2^{53 − 52}. So loss of precision first occurs with 2⁵³ + 1.

9007199254740992 (that's 9,007,199,254,740,992 or 2^53) with no guarantees :)

Program

#include <math.h>
#include <stdio.h>

int main(void) {
  double dbl = 0; /* I started with 9007199254000000, a little less than 2^53 */
  while (dbl + 1 != dbl) dbl++;
  printf("%.0f\n", dbl - 1);
  printf("%.0f\n", dbl);
  printf("%.0f\n", dbl + 1);
  return 0;
}

Result

9007199254740991
9007199254740992
9007199254740992