What are the differences between a HashMap and a Hashtable in Java? Which is more efficient for non-threaded applications? There are several differences between HashMap and Hashtable in Java: Hashtable is synchronized, whereas HashMap is not. This makes HashMap better for non-threaded applications, as unsynchronized Objects typically perform better than synchronized ones. Hashtable does
Java中的HashMap和Hashtable有什么区别？ 哪种方法对非线程应用程序更有效？ Java中的HashMap和Hashtable有几个区别： Hashtable是同步的，而HashMap不是。 这使得HashMap对于非线程应用程序更好，因为非同步对象通常比同步对象执行得更好。 Hashtable不允许null键或值。 HashMap允许一个null键和任意数量的null值。 其中一个HashMap的子类是LinkedHashMap ，所以如果您希望可预测的迭代顺序（默认为插入顺序），您
I was looking at the new APIs introduced in Android 4.2. While looking at the UserManager class I came across the following method: public boolean isUserAGoat() Used to determine whether the user making this call is subject to teleportations. Returns whether the user making this call is a goat. How and when should this be used? From their source , the method used to return false until i
我正在研究Android 4.2中引入的新API。 在查看UserManager类时，我遇到了以下方法： public boolean isUserAGoat() 用于确定发起此呼叫的用户是否需要远程传送。 返回进行此调用的用户是否是山羊。 如何和何时应该使用？ 从它们的源代码中 ，用于返回false的方法直到它在API 21中被更改。 /** * Used to determine whether the user making this call is subject to * teleportations. * @return whether the user
I'm writing some code in Java where, at some point, the flow of the program is determined by whether two int variables, "a" and "b", are non-zero (note: a and b are never negative, and never within integer overflow range). I can evaluate it with if (a != 0 && b != 0) { /* Some code */ } Or alternatively if (a*b != 0) { /* Some code */ } Because I expect that
我在Java中编写了一些代码，在某些时候，程序的流程由两个int变量“a”和“b”是否非零来确定（注意：a和b从不是负数，以及从不在整数溢出范围内）。 我可以用它评估它 if (a != 0 && b != 0) { /* Some code */ } 或者可选 if (a*b != 0) { /* Some code */ } 因为我希望每段代码运行数百万次，所以我想知道哪一个会更快。 我通过在一个巨大的随机生成的数组上进行比较来做了实验，我也很好奇看看数组的稀疏性（数据
Inspired by this question: Why is it faster to process a sorted array than an unsorted array? I wrote my own branch prediction experiment: public class BranchPrediction { public static void main(final String[] args) { long start; long sum = 0; /* No branch */ start = System.nanoTime(); sum = 0; for (long i = 0; i < 10000000000L; ++i)
受这个问题的启发：为什么处理一个有序数组比未经排序的数组更快？ 我写了自己的分支预测实验： public class BranchPrediction { public static void main(final String[] args) { long start; long sum = 0; /* No branch */ start = System.nanoTime(); sum = 0; for (long i = 0; i < 10000000000L; ++i) sum += i; System.out.println(Syst
Here's my understanding. In C programming, if I do int a then that a is created on stack and thus the memory is taken from stack. Heap plays no part here. But if I do something like int *a; a=(int*)malloc(sizeof(int)); and dynamically allocate the memory, then the reference variable will be placed on stack, but the memory it points to will be on the heap. Am I correct with my underst
这是我的理解。 在C编程中，如果我做int a那么在堆栈上创建a ，从而从栈中获取内存。 堆在这里不起作用。 但是，如果我做了类似的事情 int *a; a=(int*)malloc(sizeof(int)); 并动态分配内存，那么引用变量将被放置在堆栈上，但它指向的内存将位于堆上。 我的理解是否正确？ 现在，我拿起了关于java的这本书 无论何时你需要一个对象，你只需编写代码来使用new来创建它，并且在代码执行时将存储分配到堆上。 所以
I want to study Java again, because I leave it some years ago. Reading a book I had a problem understanding how Java allocate memory in heap and in stack. This is what I've understood - I'll try to speak about it with examples. class TestA { int a; void methodA(int b) { a = b; } int getA() { return a; } } This is a sample class to show different
我想再次学习Java，因为我几年前就离开了它。 读一本书我有一个问题，就是理解Java如何在堆栈和堆栈中分配内存。 这就是我所理解的 - 我会试着用例子来谈论它。 class TestA { int a; void methodA(int b) { a = b; } int getA() { return a; } } 这是一个示例类，以显示不同的情况。 这是我的主要： int b = 3; TestA obj = new TestA(); obj.methodA(b); obj.getA(); 那么会发
If you have java.io.InputStream object, how should you process that object and produce a String ? Suppose I have an InputStream that contains text data, and I want to convert this to a String . For example, so I can write the contents of the stream to a log file. What is the easiest way to take the InputStream and convert it to a String ? public String convertStreamToString(InputStream is)
如果你有java.io.InputStream对象，你应该如何处理这个对象并产生一个String ？ 假设我有一个包含文本数据的InputStream ，并且我想将它转换为一个String 。 例如，我可以将流的内容写入日志文件。 采用InputStream并将其转换为String的最简单方法是什么？ public String convertStreamToString(InputStream is) { // ??? } 一个很好的方法是使用Apache commons IOUtils将InputStream复制到StringWriter ...类似于 S
I use object != null a lot to avoid NullPointerException . Is there a good alternative to this? For example: if (someobject != null) { someobject.doCalc(); } This avoids a NullPointerException , when it is unknown if the object is null or not. Note that the accepted answer may be out of date, see https://stackoverflow.com/a/2386013/12943 for a more recent approach. This to me sounds
我使用object != null来避免NullPointerException 。 有没有一个很好的选择呢？ 例如： if (someobject != null) { someobject.doCalc(); } 这可以避免NullPointerException ，如果该对象为null或未知，则不会发生。 请注意，接受的答案可能已过时，请参阅https://stackoverflow.com/a/2386013/12943以获取更新的方法。 对我而言，这听起来像是一个相当普遍的问题，初级到中级开发人员往往会在某些时候面临这样的
Until today, I thought that for example: i += j; is just a shortcut for: i = i + j; But what if we try this: int i = 5; long j = 8; Then i = i + j; will not compile but i += j; will compile fine. Does it mean that in fact i += j; is a shortcut for something like this i = (type of i) (i + j) ? As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound As
直到今天，我认为，例如： i += j; 只是一个捷径： i = i + j; 但是，如果我们尝试这样做： int i = 5; long j = 8; 然后i = i + j; 不会编译，但i += j; 将编译好。 这是否意味着实际上i += j; 是这样的东西的捷径i = (type of i) (i + j) ？ 像这些问题一样，JLS持有答案。 在这种情况下，§15.26.2复合赋值运算符。 摘录： E1 op= E2形式的复合赋值表达式等价于E1 = (T)((E1) op (E2)) ，其中T是E1的类型，只
I always thought Java was pass-by-reference . However, I've seen a couple of blog posts (for example, this blog) that claim it isn't. I don't think I understand the distinction they're making. What is the explanation? Java is always pass-by-value . Unfortunately, they decided to call the location of an object a "reference". When we pass the value of an object, we
我一直认为Java是通过引用 。 不过，我见过一些博客文章（例如，这个博客），声称它不是。 我不认为我理解他们的区别。 什么是解释？ Java始终是按值传递的 。 不幸的是，他们决定将对象的位置称为“参考”。 当我们传递一个对象的值时，我们将引用传递给它。 这对初学者来说很混乱。 它是这样的： public static void main(String[] args) { Dog aDog = new Dog("Max"); // we pass the object to foo foo