Java 映射和原语

Java Mappings and Primitives

提问by James

I want to create a mapping that takes a Stringas the key and a primitiveas the value. I was looking at the Java docs and did not see that Primitive was a class type, or that they shared some kind of wrapping class.

我想创建一个以 aString为键和一个原语为值的映射。我正在查看 Java 文档，并没有看到 Primitive 是一种类类型，或者它们共享某种包装类。

How can I constrain the value to be a primitive?

如何将值限制为原始值？

Map<String, Primitive> map = new HashMap<String, Primitive>();

Map<String, Primitive> map = new HashMap<String, Primitive>();

回答by khachik

Java Autoboxingallows to create maps on Long, Integer, Doubleand then operate them using primitive values. For example:

Java 自动装箱允许创建映射Long, Integer, Double，然后使用原始值操作它们。例如：

java.util.HashMap<String, Integer> map = new java.util.HashMap<String, Integer>();
map.put("one", 1); // 1 is an integer, not an instance of Integer

If you want to store in one map different primitive types, you can to it by making a Map<String, Number>. Allows to store values of BigDecimal, BigInteger, Byte, Double, Float, Integer, Long, Short(and AtomicLong, AtomicInteger).

如果你想在一个映射中存储不同的原始类型，你可以通过创建一个Map<String, Number>. 允许存储BigDecimal, BigInteger, Byte, Double, Float, Integer, Long, Short(和AtomicLong, AtomicInteger) 的值。

Here is an example:

下面是一个例子：

Map<String, Number> map = new HashMap<String, Number>();

map.put("one", 1);
map.put("two", 2.0);
map.put("three", 1L);

for(String k:map.keySet()) {
  Number v = map.get(k);
  System.err.println(v + " is instance of " + v.getClass().getName() + ": " + v);
}

回答by mchr

Google for "Java Primitive Maps" and you will find some specialised types which avoid the need for autoboxing. An example of this is: https://labs.carrotsearch.com/hppc.html

谷歌搜索“Java Primitive Maps”，你会发现一些不需要自动装箱的特殊类型。一个例子是：https: //labs.carrotsearch.com/hppc.html

However, in general you should do fine with autoboxing as mentioned in other answers.

但是，通常您应该像其他答案中提到的那样使用自动装箱。

回答by jjnguy

You can do the following:

您可以执行以下操作：

Map<String, Integer> map = new HashMap<String, Integer>()

Then operations like:

然后操作如下：

map.put("One", 1);

will work. The primitive 1will get auto-boxed into an Integer. Likewise:

将工作。原语1将被自动装箱到Integer. 同样地：

int i = map.get("One");

will also work because the Integerwill get auto-unboxed into an int.

也将起作用，因为Integer将被自动拆箱到int.

Check out some documentation on autoboxing and autounboxing.

查看有关autoboxing 和 autounboxing 的一些文档。

回答by Andreas Dolk

Every primitive has a wrapper class, like java.lang.Longfor long.

每个原语都有一个包装类，比如java.lang.Longfor long。

So you can map the the wrapper class to Stringand, if you use Java 1.5+, simply put primitives to the map:

因此，您可以将包装类映射到String，如果您使用 Java 1.5+，只需将原语放入映射中：

 Map<String, Integer> map = new HashMap<String, Integer>();
 map.put("key", 10);
 int value = map.get("key");  // value is 10.

回答by John Vint

You would use their boxed counterpart.

你会使用他们的盒装对应物。

Map<String,Integer> map = new HashMap<String,Integer>();

Integer is an immutable boxed type of the primitive int. There are similar Short, Long, Double, Float and Byte boxed types.

Integer 是原始 int 的不可变盒装类型。有类似的 Short、Long、Double、Float 和 Byte 盒装类型。

回答by Peter Lawrey

If you need the value to be a primitive for performance reasons, you can use TObjectIntHashMapor similar.

如果出于性能原因需要将该值作为原语，则可以使用TObjectIntHashMap或类似方法。

e.g.

例如

TObjectIntHashMap<String> map = new TObjectIntHashMap();

map.put("key", 10);
int value = map.get("key");

One difference with Map<String, Integer> is that the values are of type int primitive rather than Integer object.

与 Map<String, Integer> 的一个区别是值的类型是 int 原始类型而不是 Integer 对象。

回答by Vladimir Ivanov

You can't have a primitive as key or value in Mapinterface. Instead you can use Wrapper classes, like Integer, Character, Booleanand so on.

您不能在Map界面中将基元作为键或值。相反，你可以用这样的包装类Integer，Character，Boolean等等。

Read more on wiki.

在wiki上阅读更多内容。