We're building a JSP web application, which runs inside the Apache Felix OSGi container (the web app itself is a OSGi Bundle). Now, we're facing the following problem: According to the JSP 2.0 Spec, TLD (taglib descriptors) no longer need to reside inside the web apps WEB-INF folder, but are loaded directly from the taglib's jar META-INF folder. This taglib jars usually reside ins
我们正在构建一个JSP Web应用程序，该应用程序在Apache Felix OSGi容器中运行（Web应用程序本身是一个OSGi包）。 现在，我们正面临以下问题： 根据JSP 2.0规范，TLD（taglib描述符）不再需要驻留在Web应用程序的WEB-INF文件夹内，而是直接从taglib的jar META-INF文件夹中加载。 这个taglib jar通常驻留在Web应用程序WEB-INF / lib文件夹中，但由于它们是OSGi包，它们由Felix加载。 在taglib的OSGi信息中，我们确实导入了所
I have a long set of comparisons to do in Java, and I'd like to know if one or more of them come out as true. The string of comparisons was long and difficult to read, so I broke it up for readability, and automatically went to use a shortcut operator |= rather than negativeValue = negativeValue || boolean negativeValue = negativeValue || boolean . boolean negativeValue = false; negativeVa
我在Java中做了一系列比较，我想知道它们中的一个或多个是否真实。 比较的字符串很长，很难阅读，所以为了便于阅读而将其分解，并自动使用快捷键运算符|=而不是negativeValue = negativeValue || boolean negativeValue = negativeValue || boolean 。 boolean negativeValue = false; negativeValue |= (defaultStock < 0); negativeValue |= (defaultWholesale < 0); negativeValue |= (defaultRetail < 0); negati
I'm attempting to compile Java 1.4 code that was created by IBM's WSDL2Java on Java5 without recreating the stubs and saw this error in Eclipse. I'm under the assumption that the stubs created should just compile as long as the runtime jars are available (they are). Access restriction: The type QName is not accessible due to restriction on required library C:Program FilesJavajdk1.5
我试图编译由Java的WSDL2Java创建的Java 1.4代码，而无需重新创建存根，并在Eclipse中看到此错误。 我假设创建的存根应该只在运行时罐子可用时才进行编译（它们是）。 Access restriction: The type QName is not accessible due to restriction on required library C:Program FilesJavajdk1.5.0_16jrelibrt.jar 完整的类名是javax.xml.namespace.QName 这里到底发生了什么？ 这是我试图从香肠重构猪的情况吗？ 我最
考虑到这个代码，我可以绝对确定finally块总是执行，不管什么something()是什么？ try { something(); return success; } catch (Exception e) { return failure; } finally { System.out.println("i don't know if this will get printed out."); } Yes, finally will be called. The only times finally won't be called are: If you invoke System.exit() ; If the JVM crashes fi
考虑到这个代码，我可以绝对确定finally块总是执行，不管什么something()是什么？ try { something(); return success; } catch (Exception e) { return failure; } finally { System.out.println("i don't know if this will get printed out."); } 是的， finally会被召唤。 finally不会被调用的唯一时间是： 如果你调用System.exit() ; 如果JVM首先崩溃; 如果JVM在try或catch块中达
This question already has an answer here: Why don't Java's +=, -=, *=, /= compound assignment operators require casting? 11 answers As you've now mentioned casting... there is a difference in this case: byte a = 5; a += 10; // Valid a = a + 10; // Invalid, as the expression "a + 10" is of type int From the Java Language Specification section 15.26.2: A compound assignment exp
这个问题在这里已经有了答案： 为什么Java的+ =， - =，* =，/ =复合赋值操作符需要转换？ 11个答案 正如你现在提到的那样......在这种情况下有所不同： byte a = 5; a += 10; // Valid a = a + 10; // Invalid, as the expression "a + 10" is of type int 从Java语言规范第15.26.2节： 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，只是E1只计算一次。 有趣的是，他们在规范中
What is the main difference between StringBuffer and StringBuilder ? Is there any performance issues when deciding on any one of these? StringBuffer是同步的， StringBuilder不是。 StringBuilder is faster than StringBuffer because it's not synchronized . Here's a simple benchmark test: public class Main { public static void main(String[] args) { int N = 77777777;
StringBuffer和StringBuilder的主要区别是什么？ 在决定其中的任何一个时，是否有任何性能问题？ StringBuffer是同步的， StringBuilder不是。 StringBuilder比StringBuffer快，因为它没有synchronized 。 这是一个简单的基准测试： public class Main { public static void main(String[] args) { int N = 77777777; long t; { StringBuffer sb = new StringBuffer();
This question already has an answer here: What is the “-->” operator in C++? 21 answers It's two operations. The postfix -- ( a = a - 1 but effective on the next line) and a greater than. It's equivalent to something like while (a > 0) { a = a - 1;
这个问题在这里已经有了答案： 什么是C ++中的“ - >”运算符？ 21个答案 这是两个操作。 后缀-- （ a = a - 1但在下一行有效）和大于。 这相当于类似的东西 while (a > 0) { a = a - 1;
This question already has an answer here: What is the “-->” operator in C++? 21 answers Java: Prefix/postfix of increment/decrement operators? 9 answers what confuses you is that there is no whitespace between the T-- and the > , so you might think there's a --> operator. looking like this: while (T-- > 0) { } It makes more sense, in every loop you decrease T by one
这个问题在这里已经有了答案： 什么是C ++中的“ - >”运算符？ 21个答案 Java：增量/减量运算符的前缀/后缀？ 9个答案 让你感到困惑的是T--和>之间没有空格，所以你可能会认为有一个-->运算符。 看起来像这样： while (T-- > 0) { } 更有意义的是，在每一个循环中你都将T减1 - 每次运行循环时 - （减量）运算符将从T中减去（在循环条件运行之后，因为它在T之后）。 最简单的方法是尝试一下： publi
This question already has an answer here: What is the “-->” operator in C++? 21 answers --> is not a new operator. It is just a conjunction of the operators -- and > . You first compare, and then decrement the variable. That is, i --> 0 becomes effectively i > 0; //Compare i--; //and decrement i --> 0手段i-- > 0 ，i被decrememnted和的先前值i相比0 。 --> is n
这个问题在这里已经有了答案： 什么是C ++中的“ - >”运算符？ 21个答案 -->不是新的操作员。 它只是运营商的一个联合体--和> 。 你首先比较，然后递减变量。 那是， i --> 0 变得有效 i > 0; //Compare i--; //and decrement i --> 0手段i-- > 0 ，i被decrememnted和的先前值i相比0 。 -->不是任何运营商。 这只是--和> 。 所以，当你写 i-->0这意味着比较i的价值，然后减少
This question already has an answer here: What is the “-->” operator in C++? 21 answers <-- is not a new Java operator (even though it may look like it), but there are 2 normal operators: < and -- while (var2 <-- var1) is the same as while(var2 < (--var1)) , which can be translated to plain english as: decrement the var1 variable ( --var is a prefix decrementation, ie. dec
这个问题在这里已经有了答案： 什么是C ++中的“ - >”运算符？ 21个答案 <--不是一个新的Java运算符（尽管它可能看起来像这样），但有2个正常的运算符： <和-- while (var2 <-- var1)与while(var2 < (--var1)) while (var2 <-- var1)相同，可以将其翻译为纯英文，如下所示： 递减var1变量（ --var是前缀递减，即在条件验证之前递减变量） 验证条件var2 < var1 <--在java中没有这样的操作符