Which of the following is better practice in Java 8? Java 8: joins.forEach(join -> mIrc.join(mSession, join)); Java 7: for (String join : joins) { mIrc.join(mSession, join); } I have lots of for loops that could be "simplified" with lambdas, but is there really any advantage of using them including performance and readability? EDIT I'll also extend this question to
以下哪个更适合于Java 8？ Java 8： joins.forEach(join -> mIrc.join(mSession, join)); Java 7： for (String join : joins) { mIrc.join(mSession, join); } 我有很多for循环可以用lambdas“简化”，但是真的有使用它们的优点，包括性能和可读性吗？ 编辑 我也将这个问题扩展到更长的方法 - 我知道你不能从lambda返回或破坏父函数，应该提及它们是否被比较，但是还有什么需要考虑的？ 当这些操作可以并行执
I may be teaching a "Java crash-course" soon. While it is probably safe to assume that the audience members will know Big-O notation, it is probably not safe to assume that they will know what the order of the various operations on various collection implementations is. I could take time to generate a summary matrix myself, but if it's already out there in the public domain somew
我可能很快会教授一个“Java碰撞课程”。 虽然假设观众成员可能知道Big-O符号可能是安全的，但假定他们知道各种集合实现上的各种操作的顺序可能是不安全的。 我可以花时间自行生成摘要矩阵，但如果它已经在公共领域的某个地方出现，我肯定会重用它（当然有适当的信用）。 任何人有任何指针？ 这个网站很不错，但并不特定于Java：http://bigocheatsheet.com/ Java泛型和集合这本书有这些信息（页面：188,211,222,240）。
so far I have seen two approaches of setting a variable's value in Java. Sometimes a constructor with arguments is used, others setter methods are used to set the value of each variable. I know that a constructor initialises an instance variable inside a class once a class is instantiated using the "new" Keyword. But when do we use constructors and when do we use setters? You
到目前为止，我已经看到了两种在Java中设置变量值的方法。 有时使用带参数的构造函数，其他的setter方法用于设置每个变量的值。 我知道一旦使用“new”关键字实例化了一个类，构造函数就会在类中初始化一个实例变量。 但是我们什么时候使用构造函数，什么时候使用setters？ 当您想要创建对象的新实例时，您应该使用构造函数方法，并且已经填充了这些值（一个随时可用的对象，其值已填充）。 这样你就不需要明确坐下来调用
I am very new for Android development, and I am trying to use HashMap in Android sample project. Now, am doing sample project for learn android. I just store keys and values in HashMap, i want to show the keys and their values in EditView. I followed below code in my sample project. But, first key and value only printing in EditView. Map<String, String> map = new HashMap<String,S
我对Android开发非常新，并且我正在尝试在Android示例项目中使用HashMap。 现在，我正在做学习android的示例项目。 我只是将键和值存储在HashMap中，我想在EditView中显示键和它们的值。 我在我的示例项目中遵循以下代码。 但是，第一个键和值只能在EditView中打印。 Map<String, String> map = new HashMap<String,String>(); map.put("iOS", "100"); map.put("Android", "101"); map.put("Java",
Sample URL: ../search/?attr1=value1&attr2=value2&attr4=value4 I do not know the names of attr1, att2, and attr4. I would like to be able to do something like that (or similar, don't care, just as long as I have access to the Map of request param name -> value: @RequestMapping(value = "/search/{parameters}", method = RequestMethod.GET) public void search(HttpServletRequest requ
示例网址： ../search/?attr1=value1&attr2=value2&attr4=value4 我不知道attr1，att2和attr4的名字。 我希望能够做到这样的事情（或类似的，不在乎，只要我有权访问请求参数名称的映射 - >值： @RequestMapping(value = "/search/{parameters}", method = RequestMethod.GET) public void search(HttpServletRequest request, @PathVariable Map<String,String> allRequestParams, ModelMap model) throw
This question already has an answer here: How to efficiently iterate over each entry in a 'Map'? 38 answers 我知道我对那个人有点晚了，但我会分享我的工作，以防别人帮助别人： HashMap<String, HashMap> selects = new HashMap<String, HashMap>(); for(Map.Entry<String, HashMap> entry : selects.entrySet()) { String key = entry.getKey(); HashMap value = entry.getValue
这个问题在这里已经有了答案： 如何有效地迭代'Map'中的每个条目？ 38个答案 我知道我对那个人有点晚了，但我会分享我的工作，以防别人帮助别人： HashMap<String, HashMap> selects = new HashMap<String, HashMap>(); for(Map.Entry<String, HashMap> entry : selects.entrySet()) { String key = entry.getKey(); HashMap value = entry.getValue(); // do what you have to do h
This question already has an answer here: Iterate through a HashMap [duplicate] 7 answers You're throwing away the results from your iterator. Keep the entry and you can access the key and value for printing: Iterator<Map.Entry<String, String>> i = emailDetailTree.entrySet().iterator(); while(i.hasNext()) { Map.Entry<String, String> entry = i.next(); ...
这个问题在这里已经有了答案： 迭代通过HashMap [复制] 7个答案 你抛弃了迭代器的结果。 保持输入，您可以访问打印的键和值： Iterator<Map.Entry<String, String>> i = emailDetailTree.entrySet().iterator(); while(i.hasNext()) { Map.Entry<String, String> entry = i.next(); ...
This question already has an answer here: Iterate through a HashMap [duplicate] 7 answers Here is the solution. "trying to find the most frequent word that comes up" part is not clear to me Map<Integer, Integer> map = new HashMap<Integer, Integer>(); for (Map.Entry<Integer, Integer> entry : map.entrySet()) { System.out.println("Key = " + entry.getKey() + ", V
这个问题在这里已经有了答案： 迭代通过HashMap [复制] 7个答案 这是解决方案。 “试图找出最常出现的单词”部分对我来说并不清楚 Map<Integer, Integer> map = new HashMap<Integer, Integer>(); for (Map.Entry<Integer, Integer> entry : map.entrySet()) { System.out.println("Key = " + entry.getKey() + ", Value = " + entry.getValue()); } “没有理解”试图找出最常出现的词“。你可以试试这个
This question already has an answer here: Iterate through a HashMap [duplicate] 7 answers change areaMap.get(key); to areaMap.get(lKey); I do think you should iterate using a entry set so that you could get the value easier though for (Map.Entry<String, List<X2Y2>> kv : areaMap.entrySet()) { System.out.println(kv.getValue()); }
这个问题在这里已经有了答案： 迭代通过HashMap [复制] 7个答案 改变areaMap.get(key); 到areaMap.get(lKey); 我认为你应该使用一个入口集进行迭代，这样你可以更容易地获得值 for (Map.Entry<String, List<X2Y2>> kv : areaMap.entrySet()) { System.out.println(kv.getValue()); }
This question already has an answer here: Iterate through a HashMap [duplicate] 7 answers int total =teamPlayers.get(TeamType.TEAM_ONE).size() + teamPlayers.get(TeamType.TEAM_TWO).size() should work. Instead of referencing your old post can you post your current code? hard to know what changes youve made and what suggestions you took from the last post. updated answer for iterating throu
这个问题在这里已经有了答案： 迭代通过HashMap [复制] 7个答案 int total =teamPlayers.get(TeamType.TEAM_ONE).size() + teamPlayers.get(TeamType.TEAM_TWO).size() 应该管用。 您可以发布当前的代码，而不是引用旧的帖子？ 很难知道你做了什么改变，以及你从最后一篇文章中提出的建议。 更新了您的TeamType枚举中每个条目的迭代答案： int total = 0; for(TeamType type : TeamType.values()){ total += teamPl