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As it states from oracle

Reference from Oracle Docs

Widening Primitive Conversion 19 specific conversions on primitive types are called the widening primitive conversions:

  1. byte to short, int, long, float, or double
  2. short to int, long, float, or double
  3. char to int, long, float, or double
  4. int to long, float, or double
  5. long to float or double?
  6. float to double

If a float has 32 bits and a long has 64 how is that considered widening? Shouldn't this be considered narrowing?

Bohemian
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stackoverflow
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  • i think its related with precision. I can only think about that :) – kaysush Mar 05 '13 at 16:42
  • You said "*If a long has 32 bits and a long has 64*"... make your mind up – Bohemian Mar 05 '13 at 16:44
  • I didn't understand your question : If a long has 32 bits and a long has 64 how is that considered widening? Shouldn't this be considered narrowing? The long data type is a 64-bit signed two's complement integer in java. please refer to the following link : http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html – GKP Mar 05 '13 at 16:43
  • Sorry I had a typo. See the revised question – stackoverflow Mar 05 '13 at 16:46
  • Most valuable question. Indirectly I got a solution for a long running issue. Thank you – REMITH Nov 18 '20 at 06:55

4 Answers4

27

The range of values that can be represented by a float or double is much larger than the range that can be represented by a long. Although one might lose significant digits when converting from a long to a float, it is still a "widening" operation because the range is wider.

From the Java Language Specification, §5.1.2:

A widening conversion of an int or a long value to float, or of a long value to double, may result in loss of precision - that is, the result may lose some of the least significant bits of the value. In this case, the resulting floating-point value will be a correctly rounded version of the integer value, using IEEE 754 round-to-nearest mode (§4.2.4).

Note that a double can exactly represent every possible int value.

Ted Hopp
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    @Mrshll187 - Actually, a `float` is 32 bits, of which 8 are for the exponent. There are only 24 bits available to represent the mantissa. I don't know what "I can be up to 64 bits" would mean. A float can represent some of the numbers in the range 1.40129846432481707e-45 to 3.40282346638528860e+38 (as well as a few special values like NaN and ±infinity). A `long` can represent all numbers in the range -9,223,372,036,854,775,808 to +9,223,372,036,854,775,807. You can think of each type name as the name of a subset of numbers. The subsets overlap, but each subset has elements not in the other. – Ted Hopp Mar 05 '13 at 17:04
  • Thank You Ted! I was having a very hard time grasping this as I was thinking in terms of bits. – Andrew Tobey Feb 18 '15 at 21:42
5

It is considered widening because the numbers that can be represented by a float is larger than numbers that can represented by long. Just because float uses 32 bit precision does not mean the numbers it can represent are limited to 2^32.

For instance the float (float)Long.MAX_VALUE+(float)Long.MAX_VALUE is larger than Long.MAX_VALUE, even though the float has less precision that the long.

RudolphEst
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It's considered widening because float and double can represent larger values than long. You may lose precision, but it will be possible to represent the value (at least approximately).

class stacker
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1

If you look at the matter in simple terms, it is about how data has been represented by original designers.

ideally bit depth of long(64) is larger than float(32). But float data has represented using scientific notion which allows to represent considerably much larger range

Ex: 300[original number] : 3×102 [scientific representation]

Long : -2^63 to 2^63-1

Float : (-3.4)*10^38 to (3.4)*10^38

Notice the Long(power of two) Vs Float(power of ten) representational difference here which allow float to have higher range

hope this is helpful

shanika yrs
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