java：HashMap<String, int> 不工作
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java: HashMap<String, int> not working
提问by gklots
HashMap<String, int>doesn't seem to work but HashMap<String, Integer>does work.
Any ideas why?
HashMap<String, int>似乎不起作用，但HashMap<String, Integer>确实有效。任何想法为什么？
采纳答案by cletus
You can't use primitive types as generic arguments in Java. Use instead:
在 Java 中不能使用原始类型作为泛型参数。改用：
Map<String, Integer> myMap = new HashMap<String, Integer>();
With auto-boxing/unboxingthere is little difference in the code. Auto-boxing means you can write:
使用自动装箱/拆箱，代码几乎没有区别。自动装箱意味着您可以编写：
myMap.put("foo", 3);
instead of:
代替：
myMap.put("foo", new Integer(3));
Auto-boxing means the first version is implicitly converted to the second. Auto-unboxing means you can write:
自动装箱意味着第一个版本被隐式转换为第二个版本。自动拆箱意味着您可以编写：
int i = myMap.get("foo");
instead of:
代替：
int i = myMap.get("foo").intValue();
The implicit call to intValue()means if the key isn't found it will generate a NullPointerException, for example:
隐式调用intValue()意味着如果找不到密钥，它将生成一个NullPointerException，例如：
int i = myMap.get("bar"); // NullPointerException
The reason is type erasure. Unlike, say, in C# generic types aren't retained at runtime. They are just "syntactic sugar" for explicit casting to save you doing this:
原因是类型擦除。与 C# 中的泛型类型不同，在运行时不保留。它们只是显式转换的“语法糖”，以节省您执行此操作：
Integer i = (Integer)myMap.get("foo");
To give you an example, this code is perfectly legal:
举个例子，这段代码是完全合法的：
Map<String, Integer> myMap = new HashMap<String, Integer>();
Map<Integer, String> map2 = (Map<Integer, String>)myMap;
map2.put(3, "foo");
回答by Peter Lawrey
GNU Trove support this but not using generics. http://trove4j.sourceforge.net/javadocs/gnu/trove/TObjectIntHashMap.html
GNU Trove 支持这一点，但不使用泛型。http://trove4j.sourceforge.net/javadocs/gnu/trove/TObjectIntHashMap.html
回答by Chitrapal Singh
You can use reference type in generic arguments, not primitive type. So here you should use
您可以在泛型参数中使用引用类型，而不是原始类型。所以在这里你应该使用
Map<String, Integer> myMap = new HashMap<String, Integer>();
and store value as
并将值存储为
myMap.put("abc", 5);
回答by Chitrapal Singh
You cannot use primitive types in HashMap. int, or doubledon't work. You have to use its enclosing type. for an example
您不能在HashMap. int，或者double不工作。你必须使用它的封闭类型。举个例子
Map<String,Integer> m = new HashMap<String,Integer>();
Now both are objects, so this will work.
现在两者都是对象，所以这将起作用。
回答by Younes Meridji
int is a primitive type, you can read what does mean a primitive type in java here, and a Map is an interface that has to objects as input:
int 是一种原始类型，您可以在这里阅读 java 中的原始类型的含义，而 Map 是一个接口，必须将对象作为输入：
public interface Map<K extends Object, V extends Object>
object means a class, and it means also that you can create an other class that exends from it, but you can not create a class that exends from int. So you can not use int variable as an object. I have tow solutions for your problem:
object 意味着一个类，这也意味着您可以创建一个从它扩展的其他类，但您不能创建一个从 int 扩展的类。所以你不能使用 int 变量作为对象。对于您的问题，我有两个解决方案：
Map<String, Integer> map = new HashMap<>();
or
或者
Map<String, int[]> map = new HashMap<>();
int x = 1;
//put x in map
int[] x_ = new int[]{x};
map.put("x", x_);
//get the value of x
int y = map.get("x")[0];
