Convert int to float and back doesn't give initial value

Question

This code print -46 when we cast a float to int,

This is because the information was lost during the conversion from type int to type float.

int i = 1_234_567_890;
float f = i;
System.out.println(i - (int)f);   //print  -46
System.out.println(i - f);        //print 0.0

How can one know about loss of precision while using float and double?

How can we know the result when we analyze the code before testing it ?

you can use `Math.ulp()` to get an idea of what error to expect (e.g. `Math.ulp((float) 1_234_567_890) = 128.0`) - otherwise you must understand the IEEE-754 standard — user85421, Nov 13 '19 at 14:04

David Pérez Cabrera · Accepted Answer · 2019-11-13T15:02:25.243

The float representation is quite different from the integer representation.

Floating-Point representation uses sign, exponent and mantissa and float has

With this, the maximun value that float can represent is: 340,282,346,638,528,860,000,000,000,000,000,000,000.000000

The maximun value that long can represent is 9,223,372,036,854,775,807

The maximun value that int can represent is: 2,147,483,647

Therefore, during the conversion from float to long or integer, the lost information can be of many orders of magnitude.

During the integer to float conversion you can lose up to 64u. of detail,It depends on the value. For example:

int i = 1_073_741_888;
float f = i;
System.out.println(i - (int)f);   //print  64
System.out.println(i - f);        //print 0.0

And

int i = 1_073_742_016;
float f = i;
System.out.println(i - (int)f);   //print  -64
System.out.println(i - f);        //print 0.0

The binary-integer representation of this values are:

The least significant bits are rounded when the value is greather than 16777216 (2^24 because of float's mantissa size -23 bits-).

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