问题
Just trying to understand auto-boxing, which I do apart from one thing:
Short s = 250;
Long l = 250;
The assignment to Long l fails. This, I expect, is because you cannot widen then box (i.e. it tries to widen the int value 250 to a long and then box it which it cannot do).
However, the assignment to Short s works. What is going on to make this fine? My assumption was it is still doing boxing and some kind of conversion. But if it's a case of it knowing 250 fits into a short, why does it not know that 250 will fit into a long?
回答1:
Normally, you cannot apply multiple (implicit) conversions in assignment (JLS §5.2 Assignment Conversion):
Assignment conversion occurs when the value of an expression is assigned (§15.26) to a variable: the type of the expression must be converted to the type of the variable. Assignment contexts allow the use of one of the following:
- an identity conversion (§5.1.1)
- a widening primitive conversion (§5.1.2)
- a widening reference conversion (§5.1.5)
- a boxing conversion (§5.1.7) optionally followed by a widening reference conversion
- an unboxing conversion (§5.1.8) optionally followed by a widening primitive conversion.
Long l=250; requires two conversions (widening primitive conversion followed by boxing conversion), that's why it doesn't compile.
Long l=250l; compiles because it requires a single boxing conversion.
But narrowing conversion of a constant expression is a special case, that's why Short s=250; compiles:
In addition, if the expression is a constant expression (§15.28) of type byte, short, char or int :
- A narrowing primitive conversion may be used if the type of the variable is byte, short, or char, and the value of the constant expression is representable in the type of the variable.
- A narrowing primitive conversion followed by a boxing conversion may be used if the type of the variable is :
- Byte and the value of the constant expression is representable in the type byte.
- Short and the value of the constant expression is representable in the type short.
- Character and the value of the constant expression is representable in the type char.
回答2:
Ideally, no auto narrowing should be allowed.
But since there are no byte/short literals, we can't write
byte b = 0b;
and it seems silly to
byte b = (byte)0;
so auto narrowing of constant integer is allowed so we can write
byte b = 0;
which is carried over to autoboxing case.
For long/Long, since there are long literals, this is less of a problem. Still, it should be allowed, since auto widening of signed integer is always safe.
来源：https://stackoverflow.com/questions/14425606/wrapper-classes-why-integer-literals-fail-for-long-but-work-for-anything-small