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, and it's represented exactly.
What you might want to know instead is 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 it's 253 (and -253 on the negative side):
- 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 2exponent − 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 × 253 − 52. So loss of precision first occurs with 253 + 1.
Answer from Steve Jessop on Stack OverflowThe 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, and it's represented exactly.
What you might want to know instead is 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 it's 253 (and -253 on the negative side):
- 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 2exponent − 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 × 253 − 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
Largest integer a long double can hold?
c++ - Range of decimal, double, integer, etc. data types? - Stack Overflow
What are the actual ranges of floating point and double data types in C#? - Stack Overflow
c - Representing integers in doubles - Stack Overflow
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, and it's represented exactly.
What you might want to know instead is 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 it's 253 (and -253 on the negative side):
- 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 2exponent − 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 × 253 − 52. So loss of precision first occurs with 253 + 1.
Answer from Steve Jessop on Stack OverflowAlright so this might be a stupid question but i need some help. I'm learning how to write in C++ for a class that I'm taking and I need to validate the data that a user could put in. I need to make sure it can take any number (decimal or integer) but not any string or other char. If this might work, I just want to know how large of an integer can be put into a long double. My idea is to use a loop to get it work and I want to use a long double for the decision statement so that the user could input valid info if they put something wrong. If there is a better way to do this, please let me know, but I am still only learning out so I ask that you're patient with me please.
Edit: Thank you all so much for the help, I'm only 3 weeks into this course and technically should not have learned loops or if else statements yet. I'm a little confused on the answers but Ill continue to try and figure them out. What I am specifically looking for is which what is the largest form of whole number that can be stored in a float that can also use any and all numbers before it. Like if 1 million was that largest a float could hold for integers, and I want to use 999,999 or 999,998 etc. etc. I'm using code blocks for anyone wondering since I've seen comments saying that the answer might vary depending on what I'm coding on
The ranges are actually –infinity to +infinity.
The largest finite float is 340282346638528859811704183484516925440. This is 2128–2128–24.
The largest finite double is 179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368. This is 21024–21024–53.
The ranges are represented in "exponential format" for conciseness. For example, +1.7e+308 means 17 followed by 307 zeros:
1,700,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
So the exponential format is preferred for such large numbers. And the same goes for extremely small numbers.
Also, take a look at this reading by Jon Skeet.
An IEEE754 64-bit double can represent any 32-bit integer, simply because it has 53-odd(a) bits available for precision and the 32-bit integer only needs, well, 32 :-)
It would be plausible for a (non IEEE754 double precision) 64-bit floating point number to have less than 32 bits of precision. That would allow truly huge numbers (due to the exponent) but at the cost of precision.
The bottom line is that, provided there are more bits of precision in the mantissa of the floating point number than there are in the integer (and enough bits in the exponent to scale it), then it can be represented without loss of precision.
(a) Technically, the 53rd bit of precision is an implied 1 at the start of the sequence so the amount of "variablity" may only be 52 bits. Whether it's 52 or 53, it's still enough bits to represent every 32-bit integer.
Yes. A float (or double) is guaranteed to exactly represent any integer that does not need to be truncated. For a double, there is 53 bits of precision, so that is more than enough to exactly represent any 32 bit integer, and a tiny (statistically speaking) proportion of 64 bit ones too.