Implicit narrowing conversion

[Gyanesh Sharma]

The implic narrowing conversion using the compound operators are equivalant to explicit conversion. I would like to share the following code sample:

class Test {
public static voice main(String[] args) {
byte b = Byte.MAX_VALUE;
byte b +=1; // compiles fine
System.out.println(b); // prints -128
byte b = b + 1; // compiler error.
}
}

[Lakshmanan Arunachalam]

b += 1;
implicitly means,
b = (Type of b) (b + 1);

Thats why it compiled without error.

b = b + 1;

In the above operation , b + 1 results int result, which can't be assigned to a byte.

Hope this helps !

[Gyanesh Sharma]

Yep... I just wanted to share it with you guys, because we often talk about the implict widening conversion. Hope this helps other folks who are also preparing for the exam.

[manogna edintipal]

I expect the last line would give compile time error

but why its compiling fine???

 
int i=Integer.MAX_VALUE;
    i+=1;
    i=i+1;
 

[Gyanesh Sharma]

You are right, the last line would give a compiler error. Notice the comment after the line.

[Ramakrishna RangaRaoOne AllamOne]

i = i+1 --- since both the operands are int , the result will be int .
b = b+1 -- since both the operands are less than int , the result will be int .
in a binary operator , if both the operands are less than int , then the result will become int . if one of the operands is wider than int , then the result will be WIDER of the two .
byte+short also results in int .
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Hope its clear .