在Java中长时间取消装箱

Question

In some code I see this:

private void compute(Long a, Long b, Long c) {
        long result = a-(b+c);
...

It seems a bit strange that the result is stored in a primitive long instead of a Long object corresponding to its operands.

Are there any reason that a result should be stored as a primitive?

Answer 1

It seems a bit strange that the result is stored in a primitive long instead of a Long object corresponding to its operands.

No, what is "strange" is that you can use the + and - operators on Long objects. Before Java 5, this would have been a syntax error. Then autoboxing/unboxing was introduced. What you're seeing in this code is autounboxing : the operators require primtives, so the compiler automatically inserts a call to longValue() on the objects. The arithmetic is then performed on primitive long values, and the result is also a long that can be stored without further conversion on the variable.

As for why the code does this, the real question is why someone would use the Long type instead of long . Possible reasons:

• The values come from some library/API that delivers Long values.
• The values are stored in collections ( List , Map ), which cannot hold primitives.
• Sloppiness or cargo cult programming .
• The ability to have null values is required, eg to signal unavailable or uninitialized data.

Note that the ability of Long to hold null values means that the calculation (or more specifically, the longValue() calls inserted by the compiler) can fail with a NullPointerException - a possibility the code should deal with somehow.

Answer 2

原因很明显：result被声明为原始。

Answer 3

The arithmetic operators + and - are not defined for boxed types (eg Long) but for primitive types (eg long).

The result is also a long. See Autoboxing and Unboxing tutorial

Autoboxing this into a Long would result in a small performance cost. It is also unnecessary because

1. We know it will be non-null (if a, b or c were null, a NullPointerException would occur).
2. It would be autoboxed implicitly if we use it later where a Long is required.

Answer 4

Based on your needs.I mean the decelaration.

Autoboxing and unboxing can happen anywhere where an object is expected and primitive type is available

Answer 5

Usually you should prefer using primitives, especially if you are certain they cannot be null. If you insist on using the boxed types always think extra hard about what happens when it is null. Java will do the boxing and unboxing automatically for you, but staring at an int and wondering why you got a NullPointerException can be fun.

Answer 6

从Java 1.5开始，自动装箱和拆箱会在需要时隐式发生。

Answer 7

The following line:

long result = a-(b+c);

...asks Java to take the result of the expression using 3 Long s, and then store it in a primitive long. Before Java 5, it would complain about the types not matching - but these days it just assumes you mean what you say and automatically does the conversion from object to primitive type for you.

In this example however, unless there's some other good reason not presented here, there's absolutely no point having the parameters as the boxed, object type in the first place.

Answer 8

As per the javadoc

Boxing conversion converts expressions of primitive 
type to corresponding expressions of reference type. 
Specifically, the following nine conversions are called the boxing conversions:

From type boolean to type Boolean

From type byte to type Byte

From type short to type Short

From type char to type Character

From type int to type Integer

From type long to type Long

From type float to type Float

From type double to type Double

From the null type to the null type


Ideally, boxing a given primitive value p, would always yield an identical reference.     
In practice, this may not be feasible using existing implementation techniques. The  
rules above are a pragmatic compromise. The final clause above requires that certain 
common values always be boxed into indistinguishable objects. The implementation may  
cache these, lazily or eagerly. For other values, this formulation disallows any 
assumptions about the identity of the boxed values on the programmer's part. This would 
allow (but not require) sharing of some or all of these references.

This ensures that in most common cases, the behavior will be the desired one, without     
imposing an undue performance penalty, especially on small devices. Less memory-limited 
implementations might, for example, cache all char and short values, as well as int and 
long values in the range of -32K to +32K.`

Here is the Oracle Doc source

Answer 9

The answer for your doubt is

1. autoboxing and autounboxing in Java which converts from primitive to wrapper class objects and vice versa respectively.
2. autoboxing means internally compiler uses valueOf() method of primitive classes ans autounboxing means internally compiler uses xxxValue() method.
3. Suppose for private void compute(Long a, Long b, Long c) { long result = a-(b+c);

its makes this conversion a.longValue()-(b.longValue()+c.longValue()) Which means even before your statement performs addition the compiler provides the primitives of long type as input to your operands Remember that this goes in hand as JAVA is statically and strongly typed language.

Hence you get long type output

I hope i cleared your doubt