CODEWITHSUNDEEP
javaarrayscastingtype-conversion
Figitaki
Figitaki

Reputation: 139

Add bytes with type casting, Java

I am trying to add two values in a byte array. This is my code:

byte[] ars = {3,6,9,2,4};
ars[0] = (byte)ars[0] + (byte)ars[4];
System.out.println( ars[0] );

I get this error on compilation:

Main.java:9: possible loss of precision
found   : int
required: byte
    ars[0] = (byte)ars[0] + (byte)ars[4];
                          ^
1 error

Any help is, as always, much appreciated.

Upvotes: 12

Views: 22176

Answers (7)

edthethird
edthethird

Reputation: 6263

close, but a little off.

ars[0] = (byte)(ars[0] + ars[4]);

keep in mind ars[0] and ars[4] are already bytes, so there is no need to cast them to bytes.

Instead, cast the result of the summation to a byte.

Upvotes: 15

Abhishek Choudhary
Abhishek Choudhary

Reputation: 8385

ars[0] = (byte)(ars[0] + ars[4]);

Upvotes: 2

Abhishek Agarwal
Abhishek Agarwal

Reputation: 51

I cam across this question a while back, and collected all the findings here : http://downwithjava.wordpress.com/2012/11/01/explanation-to-teaser-2/

We all know bytes get converted to ints during an arithmetic operation. But why does this happen? Because JVM has no arithmetic instructions defined for bytes. byte type variables have to be added by first 'numerically promoting' them to 'int' type, and then adding. Why are there no arithmetic instructions for the byte type in JVM? The JVM spec clearly says:

The Java virtual machine provides the most direct support for data of type int. This is partly in anticipation of efficient implementations of the Java virtual machine's operand stacks and local variable arrays. It is also motivated by the frequency of int data in typical programs. Other integral types have less direct support. There are no byte, char, or short versions of the store, load, or add instructions, for instance.

Upvotes: 5

flyinrhyno
flyinrhyno

Reputation: 159

public static void main(String args[]){
    byte[] ars = {3,6,9,2,4};
    ars[0] = (byte)(ars[0] + ars[4]);
    System.out.println( ars[0] );
}

this happens for the reason that java automatically converts expressions which use byte and short variables to int... this is to avoid potential risk of overflow.... as a result even if result may be in the range of byte java promotes type of expression to int

Upvotes: 4

gmhk
gmhk

Reputation: 15940

byte[] ars = {3,6,9,2,4};
        ars[0] = (byte) (ars[0] + ars[4]);
        System.out.println( ars[0] );

This will work, try it

Upvotes: 2

Taymon
Taymon

Reputation: 25666

In Java, the sum of two bytes is an int. This is because, for instance, two numbers under 127 can add to a number over 127, and by default Java uses ints for almost all numbers.

To satisfy the compiler, replace the line in question with this:

ars[0] = (byte)(ars[0] + ars[4]);

Upvotes: 14

Chetter Hummin
Chetter Hummin

Reputation: 6817

Please replace that line with the following

ars[0] = (byte)(ars[0]+ars[4]);

Upvotes: 2

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