CODEWITHSUNDEEP
javatype-conversionbyteaddition
Nirmit Dalal
Nirmit Dalal

Reputation: 325

Byte typecasting in java

the program gives me loss of precision error but i cant think of any precision loss since numbers are small

This is the code ##

 class Demo  
 {  
   public static void main(String args[])  
   {  
      byte b1=3;  
      byte b2=2;  
      byte b3=b1+b2;  
      System.out.println(b3);  
   }  
 }

Upvotes: 0

Views: 3222

Answers (4)

LMK
LMK

Reputation: 2952

when ever you do +,-,*,/,% java internally uses a function

ex: max(int,dataType of operand 1,dataType of operand 2);

here in your code
max(int, byte,byte) ==> which one is bigger ? ==> int is bigger

so you may get POSSIBLE LOSS OF PRESSION
found :int

required : byte

Another Example 

 short a =10;
 byte b=20;
 short c = a+b;

now:

internally: max(int,OP1,OP2);

ie max(int,short,byte) ==> which is bigger Data Type? int

so java excepts int 

 int c = a+b;  (works fine)

Another Example : 

 long a =10;
 byte b=20;
 short c = a+b;

now:

internally: max(int,OP1,OP2);

ie max(int,long,byte) ==> which is bigger Data Type? long

so java excepts long 

 long c = a+b;(works fine)

Another Example: 

byte b = 10;
b = b+1;

now,
max(int,byte)==> which is bigger Data Type ? int
so, 

int c = b+1;(works fine) or b = (byte) b+1;

hope you understand

Upvotes: 0

Jon Skeet
Jon Skeet

Reputation: 1500385

The addition expression b1 + b2 is of type int - there aren't any addition operators defined on smaller types than int. So in order to convert that back to a byte, you have to cast:

byte b3 = (byte) (b1 + b2);

Note that although you happen to know that the values are small, the compiler doesn't care about the values you've set in the previous two lines - they could both be 100 for all it knows. Likewise although you know that the int you're trying to assign to a byte variable is only the result of adding two byte values together (or rather, two values promoted from byte to int), the expression as a whole is just int and could have come from anywhere as far as the language is concerned.

(The fact that the addition can overflow is a separate matter, but that would be an inconsistent argument - after all, you can add two int values together and store the result in an int variable, even though the addition could have overflowed easily.)

Upvotes: 9

Ruchira Gayan Ranaweera
Ruchira Gayan Ranaweera

Reputation: 35557

    byte b1=3;
    byte b2=2;
    byte b3=b1+b2;  // you can't use byte here, Every time addition will result 
                     int value

Because, If you trying to cast an int which is larger than byte range, there is a loss part of that value. So addition will not allow to use byte here.

Upvotes: 3

RaceBase
RaceBase

Reputation: 18848

Because 127 is the last value of byte. Assume b1 = 127 and b2 = 2

now what happends b = b1+b2 = 129 [which is out of byte range, i.e. it's in int range]

now if you cast it b = (byte)(b1+b2), you will get -127 this is due to rounding of the value to byte.

Upvotes: 3

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