integer literals

[Rolf Weasel]

floating point literals default to type double. do integers default to any one of the integer types?

[paul wheaton]

They default to "int"

[Rolf Weasel]

What if they are >Integer.MAX_VALUE?

[Tony Alicea]

Then you need to use type <CODE>long</CODE>

[maha anna]

Tony!
What do you mean? Integer.MAX_VALUE is also an int right? Integer.MAX_VALUE defaults to primitive int.
regds
maha anna

[Jim Yingst]

Yes - but anything larger than Integer.MAX_VALUE can't fit in an int, so it has to be long. (See the little ">" symbol in Rolf's post? That meant "larger than Integer.MAX_VALUE".)

[maha anna]

yes Jim. I think I am reading toooooo fast.
maha anna

[Rolf Weasel]

paul, how can integer literals default to int? that would mean :
short s = 1;
would cause a compile time error without an explicit cast.

[Jim Yingst]

That's a special case. From the JLS section on assignment conversion:
<blockquote>In addition, a narrowing primitive conversion may be used if all of the following conditions are satisfied:

• The expression is a constant expression of type int.
  
• The type of the variable is byte, short, or char.
  
• The value of the expression (which is known at compile time, because it is a constant expression) is representable in the type of the variable.
</blockquote>
That's the case here, so the compiler performs a narrowing conversion without a cast.
To see that the default is int, try this:
<code><pre>public class Test {
static void method(byte b) {
System.out.println(b + " is a byte!");
}
static void method(short s) {
System.out.println(s + " is a short!");
}
static void method(int i) {
System.out.println(i + " is an int!");
}
static void method(long l) {
System.out.println(l + " is a long!");
}
public static void main(String[] args) {
method(1);
}
}</pre></code>

[Rolf Weasel]

Thanks, Jim!
I ran the code and u were right.
I also tried the following:
public class Test {
static void method(byte b) {
System.out.println(b + " is a byte!");
}
static void method(short s) {
System.out.println(s + " is a short!");
}
static void method(int i) {
System.out.println(i + " is an int!");
}
static void method(long l) {
System.out.println(l + " is a long!");
}
public static void main(String[] args) {
method( Integer.MAX_VALUE+1 );
}
}
and i got :
"-2147483648 is an int!"
should'nt Integer.MAX_VALUE+1 default to long?

[Rolf Weasel]

that's funny. i just tried:
public static void main(String[] args) {
long l = 2147483648;
method( l );
}
and i got a compile time error saying "Integer literal out of range."
So it looks like all integer literals appearing in code must be in the range -2147483648 to 2147483647.

[Rolf Weasel]

even an explicit cast did'nt work:
public static void main(String[] args) {
long l = (long) 2147483648;
method( l );
}

[Tony Alicea]

No; it makes the number "roll over" making the most significant bit a ONE, therefore a negative number.

[Rolf Weasel]

Coming back to my earlier question, how do u specify a literal of value >Integer.MAX_VALUE?

[Rolf Weasel]

Got the answer in the JLS. gotta use 2147483648L. Man, i clean forgot about L!

[Anonymous]

Is Rolf Weasel discussing with himself?