Casting of long to float

[ani pillai]

Why is the casting of long to float implicit. I would think that because long is 64 bits and float 32 bits, an explicit cast would be required. But that is not the case...any ideas??

[Henry Wong]

This question is actually commonly asked on this forum -- you should search for previous topics.

Henry

[Morteza Manavi-Parast]

Or you can read the following explanation:

A widening conversion from a long to a float value or from a long to a double value (which both have 64 bits) may result in a "loss of precision". The floating point in the target type is then a correctly rounded approximation of the integer value. Note that precision is the number of significant bits in the value and most not be confused with magnitude which relates how big a value can be represented.

Not make sense? Here is an example:

long bigInt = 98765432112345678L;
float fp = bigInt; //Widening but loss of precision: "fp= 9.8765436E16"

You should know that float and double can store much bigger values than for example long. You can prove it to yourself by printing the Float.MAX_VALUE or Double.MAX_VALUE. So they have bigger magnitude and smaller precision than long

Basically we use the term narrowing (which needs an explicit cast) whenever there is a possibility in loss of magnitude information and not solely loss precision. e.g: conversion from long to int.

Hope this helped...

[ani pillai]

Got it...went through this article http://radio.javaranch.com/corey/2004/06/01/1086120368000.html

Thanks for your help