CODEWITHSUNDEEP
java
Andrei Ciobanu
Andrei Ciobanu

Reputation: 12838

Compilation error: casting

Can somebody explain me why the following piece of code fails to compile. The error is: "Possible loss of precision." :

byte a = 50;
byte b = 40;
byte sum = (byte)a + b;
System.out.println(sum);

Thank you.

Upvotes: 2

Views: 151

Answers (4)

polygenelubricants
polygenelubricants

Reputation: 383746

You did right by identifying that a cast is required, but unfortunately you did not apply it to the right expression due to precedence of operators.

Consider the following snippet:

static void f(char ch) {
    System.out.println("f(char)");
}
static void f(int i) {
    System.out.println("f(int)");
}
public static void main(String[] args) {
    char ch = 'X';
    f( (char)  ch + 1  ); // prints "f(int)"
    f( (char) (ch + 1) ); // prints "f(char)"
}

The cast has a higher precedence than the addition, which is why the snippet prints what it does. That is, the first call is equivalent to f( ((char) ch) + 1 );. The result of the addition is an int, which is why the f(int) overload is invoked.

The lesson here is that you should always use parenthesis unless you're doing a very simple cast. In general, always consider using parentheses to make the order of evaluation explicit, even if they're not necessary. They lead to a better, more readable code.

Upvotes: 1

Jesper
Jesper

Reputation: 206796

Note that the cast has a higher precedence than the + operator. Your code does this:

byte a = 50;
byte b = 40;
byte sum = ((byte)a) + b;
System.out.println(sum);

The cast is redundant, since a is already a byte. You probably meant this:

byte a = 50;
byte b = 40;
byte sum = (byte) (a + b);
System.out.println(sum);

Upvotes: 4

user132014
user132014

Reputation:

Because the two byte variables are operands to +, they are implicitly promoted to int. This is called Numeric Promotion. Because int is larger than byte, and the result of a + b yields int, casting to byte possibly chops off some bits, as int is larger than byte. Hence the "loss of precision"

Doc for implicit numeric promotion:

http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#170983

Doc for the size of types:

http://java.sun.com/docs/books/tutorial/java/nutsandbolts/datatypes.html

Upvotes: 3

mfx
mfx

Reputation: 7388

Bytes are added using "int" arithmetic; thus the result is an int and must be cast back from int to byte, which leads to the possibility of truncation.

Upvotes: 1

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