why compilation fails here

[Divyya Joshi]

The following code is from K&B . And it explains that short needs to be cast.
I am not abe to understand this.

class Alien {
String invade(short ships) { return "a few"; }
String invade(short... ships) { return "many"; }
}
class Defender {
public static void main(String [] args) {
System.out.println(new Alien().invade(7));
} }

Kindly explain this to me

[Hama Kamal]

Divyya Joshi wrote:The following code is from K&B . And it explains that short needs to be cast.
I am not abe to understand this.

class Alien {
String invade(short ships) { return "a few"; }
String invade(short... ships) { return "many"; }
}
class Defender {
public static void main(String [] args) {
System.out.println(new Alien().invade(7));
} }

Kindly explain this to me

7 is

 int

which is bigger than

short

, so it needs to be cast

[Hama Kamal]

it works this way

class Alien {
String invade(short ships) { return "a few"; }
String invade(short... ships) { return "many"; }
}
class Defender {
public static void main(String [] args) {
System.out.println(new Alien().invade((short)7));
} }

[Divyya Joshi]

Thanks Hama

[Angelica Bunghez]

The implicit downcasting from int to short will be done only at assignment operators, not at passing arguments in the methods

[Angelica Bunghez]

and the implicit downcasting at assignments will be done only for literals in range ( 7 is a int literal in range of short primitive type) -( this is our case) and for the constants.

[Alex Theedom]

May I also add that if the value is not within the range of the short (-32,768 to 32,767 (inclusive)) an explicit cast is required and that this value must be within the range of the int (-2,147,483,648 to 2,147,483,647 (inclusive)). Note that all literal values are integers UNLESS they are denoted as floats or doubles. See the following code snippet:

	
short s1 =  32767; // short maximum value = 32767
short s2 = (short) 32768; // literal exceeds the maximum value of a short, therefore explicit cast required
short s3 = (short) 2147483647 ; // integer maximum value = 2147483647 therefore all OK
short s4 = (short) 2147483648 ; // integer maximum value = 2147483647 therefore the literal is out of range. Will not compile.
short s5 = (short) 2147483648f ; // implicit cast from float to integer then explicit cast from integer to short.
short s6 = (short) 2147483648d ; // implicit cast from double to integer then explicit cast from integer to short.	

If you print s3, s5 and s6 the result will be -1.

[Chandraprakash Sarathe]

The implicit downcasting from int to short will be done only at assignment operators, not at passing arguments in the methods

short s1 =3;
short y =  s1 +(short)4;

Here it fails to compile at line 2 which requires explicit downcasting.Means sum of two short values results an integer ..dont know why ?

[Stephan van Hulst]

Because the + operator always returns an int, unless one of the operands is a long, float, double or String. That's just how it works.

[Alex Theedom]

It is true that the + operator returns integers even if the arguments are shorts and that you must cast the result to a short. However this only applies when the operation includes a variable. Like this:

		short x = 1;
		short y = (short) (1 + x);

However in the following two circumstances it is not required to cast to short:

		short z = 1 + 4; // the two values are within the range of shorts and are implicitly cast to shorts. 

		short a = 0;
		a+=1;

Applies equally for all arithmetic operators (+, -, *, %, /) and byte, char and integers, but not floats or doubles.

[Helen Ma]

I have a question about implicit downcasting from long to int.
int a = 9 L //won't compile and implicit downcasting won't work
int a = (int) 9 L //compile
We need to do explicit downcasting.

But short a = 9 // implicity downcasting works.

Why implicit downcasting works for short, not for long?