System.out.printf("%1$c",65536);

we know the range of char is 0 to 65535..then why does this compile ?

well, the number is an integer expression. primative types don't have the elaborate boundary checking like arrays would.

so when integer this value is converted to a char internally, it is going to truncate the 17th bit, and render the remaining part that does fit into the char storage.

By that logic, i would expect these two statements to emit the same output.

primative types don't have the elaborate boundary checking like arrays would

what does this mean?

He means if you had an array of 3 objects, and tried to do array[4], it would throw an exception. With primitives, however, if you go outside their bounds they simply drop the bits that are outside their range and use what is left.

then why does char c=65536; doesnt compile?

and how does integer value is converted to char internally?

the cast is provided internally? is there any rule stating so?

Ankur kothari wrote:the cast is provided internally? is there any rule stating so?

It seems like there is a cast, but I don't know. I'm searching now, but hoping some Java guru will pop in knowing what is up.

me too.....waiting for someone...

Ankur kothari wrote:me too.....waiting for someone...

Perhaps you could clarify what your question is now. It isn't clear on what you know now.

am still looking for a rule which says about internal casting....because i know that internally casting is done

I don't think there's any such rule. I don't even think that "internal" casting is a concept in Java. Did you read that phrase somewhere?

First of all, the compiler does not know that your printf expression is expecting a char character. The method actually being invoked in printf is:

Therefore, your 65536 are autimatically converted to an Integer object thanks to autoboxing.

I got "?" from System.out.printf("%1$c",65536);
don't know why

Edwin Dalorzo wrote:First of all, the compiler does not know that your printf expression is expecting a char character. The method actually being invoked in printf is:

Therefore, your 65536 are autimatically converted to an Integer object thanks to autoboxing.

Agreed. This question has nothing to do with casting -- and whether it should compile or not. The method is implemented as varargs of objects, so there is no way for the compiler to even detect that the type is different than expected. It must be done at runtime.

Henry

zheng li wrote:I got "?" from System.out.printf("%1$c",65536);
don't know why

It is a "?" mark because there is no character font data for that unicode character. As for how printf() behaves when you pass an int to be formatted as a %c, the javadoc is generally the best resource.

http://java.sun.com/javase/6/docs/api/java/util/Formatter.html

Basically, when you pass a value that is larger than a char, it will be treated as a code point unicode.

If it is out of range, it will throw an exception. And the reason it doesn't throw an exception here, is because a code point unicode character is actually larger than 16 bits. It can span from 0 to 0x10FFFF.

Henry