2

I'm getting an error when I'm trying to do somethink like this:

byte a = 23;
a = a - 1;

The compiler gives this error: Test.java:8: possible loss of precision found : int required: byte a = a - 1; ^ 1 error

Casting doesn't solve the error... Why the compiler don't let me do it? Should I need to transform the variable 'a' into an int?

mevqz
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  • This has been asked hundred of times. See eg http://stackoverflow.com/questions/1660856/promotion-in-java http://stackoverflow.com/questions/927391/why-is-a-cast-required-for-byte-subtraction-in-c/941677 – leonbloy Mar 20 '13 at 20:02

4 Answers4

6

Do like this.

a = (byte)(a - 1);

When you subtract 1 from a then its integer value. So to get assign the result in byte you need to do explicit type casting.

Ajay S
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  • Thanks! I applied the cast in the '1' instead of the result of the expression, that was my error. Thanks again. – mevqz Mar 20 '13 at 20:08
5

In Java math, everything is promoted to at least an int before the computation. This is called Binary Numeric Promotion (JLS 5.6.2). So that's why the compiler found an int. To resolve this, cast the result of the entire expression back to byte:

a = (byte) (a - 1);
rgettman
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3

a = a - 1; // here before subtraction a is promoted to int data type and result of 'a-1' becomes int which can't be stored in byte as (byte = 8bits and int = 32 bits).

Thats why you'll have to cast it to a byte as follows : 

a = (byte) (a - 1);
VishalDevgire
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1

Do this:

a -= 1;

You even don't need explicit cast, compiler/JVM will do it for you.

Should you change the variable type to int nobody can say, having only information you provided.

A variable type is defined by the task you are planning to perform with it.

If your variable a counts fingers on someone's hands, why would you use int? Type byte is more than enough for that.

Alex Kreutznaer
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