java：HashMap< String，int>不工作 [英] java: HashMap<String, int> not working

问题描述

HashMap< String，int> 似乎不起作用，但 HashMap< String，Integer> 。
任何想法为什么？

HashMap<String, int> doesn't seem to work but HashMap<String, Integer> does work. Any ideas why?

推荐答案

Java中不能使用基本类型作为泛型参数。改为使用：

You can't use primitive types as generic arguments in Java. Use instead:

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

使用 auto-boxing / unboxing 代码没有什么区别。自动装箱意味着你可以写：

With auto-boxing/unboxing there is little difference in the code. Auto-boxing means you can write:

myMap.put("foo", 3);

而不是：

instead of:

myMap.put("foo", new Integer(3));

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：

Auto-boxing means the first version is implicitly converted to the second. Auto-unboxing means you can write:

int i = myMap.get("foo");

而不是：

instead of:

int i = myMap.get("foo").intValue();

对 intValue（）的隐式调用表示如果未找到密钥，它将生成 NullPointerException ，例如：

The implicit call to intValue() means if the key isn't found it will generate a NullPointerException, for example:

int i = myMap.get("bar"); // NullPointerException

原因是 type erasure 。与C＃不同，通用类型不会在运行时保留。它们只是语法糖，用于显式投射以节省您的工作：

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:

Integer i = (Integer)myMap.get("foo");

举个例子，这段代码是完全合法的：

To give you an example, this code is perfectly legal:

Map<String, Integer> myMap = new HashMap<String, Integer>();
Map<Integer, String> map2 = (Map<Integer, String>)myMap;
map2.put(3, "foo");

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