int a = 1L; This doesn't compile (of course). incompatible types: possible lossy conversion from long to int
int b = 0; b += Long.MAX_VALUE; This does compile!
But why is it allowed?
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2 Answers
When you do += that's a compound statement and Compiler internally casts it. Where as in first case the compiler straight way shouted at you since it is a direct statement :)
The line
b += Long.MAX_VALUE; Compiler version of it equivalent of
b += (int)Long.MAX_VALUE; of course there will be lossy conversion from conversion of long to int.
http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.26.2
A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.
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coder
ah i c, didn't found it in the spec myself. Well a warning of the compiler/IDE would be nice.
Actually the compiler is smarter than you think. At compile time itself, it will replace the actual value of the expression b+=Long.MAX_VALUE by -1. So, Long.MAX_VALUE is converted to an int and assigned to the int field at compile time itself)
Byte code :
public static void main(java.lang.String[]); descriptor: ([Ljava/lang/String;)V flags: ACC_PUBLIC, ACC_STATIC Code: stack=1, locals=2, args_size=1 0: iconst_0 1: istore_1 2: iinc 1, -1 // Here 5: return LineNumberTable: line 4: 0 line 5: 2 line 6: 5 LocalVariableTable: Start Length Slot Name Signature 0 6 0 args [Ljava/lang/String; 2 4 1 b I 
1 Comment
coder
hm currently i am thinking the compiler is too "smart" here