Type cast problem.

[wybrand]

Hai fellow programmers.

Can you please give me an answer to the following??

See next code-fragment:

class a {

public static void main( String [] arg ) {
short a;
int b;

// Implicit cast.
a = 65;

// Explicit cast which compiles.
a = (int)65;

// Explicit cast which does not compile.
b = (long)65;
}
}

The first statement compiles because of implicit typecasting and there
is no loss of information (65 < 2^(16-1)-1).
The third one does not compile because we do an explicit typecast and
a long is 64 bytes and an integer is 32.
Why does the second statement compile?? A integer doesn't fit in a
short, just like a long does not fit in a integer (third statement).
It looks like if a implicit typecast takes place just like the fist
statement.

Thanks in advance.

Greetings,

Wybrand.

 

[Joona I Palaste]

Hai fellow programmers.

Can you please give me an answer to the following??

What's wrong with the answers already given on this newsgroup?

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[Chris Smith]

(e-mail address removed) says...

short a;
int b;

// Implicit cast.
a = 65;

// Explicit cast which compiles.
a = (int)65;

// Explicit cast which does not compile.
b = (long)65;

The first statement compiles because of implicit typecasting and there
is no loss of information (65 < 2^(16-1)-1).
The third one does not compile because we do an explicit typecast and
a long is 64 bytes and an integer is 32.
Why does the second statement compile?? A integer doesn't fit in a
short, just like a long does not fit in a integer (third statement).
It looks like if a implicit typecast takes place just like the fist
statement.

The second statement is identical to the first, since the literal '65'
is already an int. The same rules for assignment conversion apply, and
so it's no less likely to compile that the first statement. These
assignment conversion rules don't apply to 'long' variables, so the
third statement fails to compile. This is in JLS section 5.2, which you
are free to read.

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