If short is automatically promoted to int in arithmetic operations, then why is:
short thirty = 10 * 3; A legal assignment to the short variable thirty?
In turn, this:
short ten = 10; short three = 3; short thirty = ten * three; // DOES NOT COMPILE AS EXPECTED as well as this:
int ten = 10; int three = 3; short thirty = ten * three; // DOES NOT COMPILE AS EXPECTED does not compile because assigning an int value to a short is not allowed without casting as expected.
Is there something special going on about numerical literals?
Because the compiler replaces 10*3 with 30 at compile time itself. So,effectively : short thirty = 10 * 3 is calculated at compile time.
Try changing ten and three to final short (making them compile time constants) and see what happens :P
Examine byte-code using javap -v for both verisions (10*3 and final short). You will be able to see that there is little difference.
Ok, So, here is the byte code difference for different cases.
Case -1 :
Java Code : main() { short s = 10*3; }
Byte code :
stack=1, locals=2, args_size=1 0: bipush 30 // directly push 30 into "s" 2: istore_1 3: return Case -2 :
public static void main(String arf[]) { final short s1= 10; final short s2 = 3; short s = s1*s2; } Byte code :
stack=1, locals=4, args_size=1 0: bipush 10 2: istore_1 3: iconst_3 4: istore_2 5: bipush 30 // AGAIN, push 30 directly into "s" 7: istore_3 8: return Case -3 :
public static void main(String arf[]) throws Exception { short s1= 10; short s2 = 3; int s = s1*s2; } Byte-code :
stack=2, locals=4, args_size=1 0: bipush 10 // push constant 10 2: istore_1 3: iconst_3 // use constant 3 4: istore_2 5: iload_1 6: iload_2 7: imul 8: istore_3 9: return In the above case, 10 and 3 are taken from the local variables s1 and s2
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