在stackoverflow上多次提到自动模块,但我找不到自动模块的完整,简洁和自给自足的定义.
那么,什么是自动模块？它是否导出所有包裹？它打开所有包裹吗？它是否读取所有其他模块？
我知道我们可以使用"var"关键字在Kotlin中定义变量:
var foo = 3
最新的java更新(java 10)也引入了"var"类型:
var bar = new int[]{1, 2, 3}; // int[] bar = {1, 2, 3}
我的问题是,这些语言之间使用"var"的区别是什么？
函数可以是协变的和逆变的.这种协变/逆变二元性也可以应用于monad吗？
就像是:
class Monad m where
return :: a -> m a
(>>=) :: m a -> (a -> m b) -> m b
class ContraMonad m where
return :: a -> m a
contrabind :: m a -> (b -> m a) -> m b
请问ContraMonad类有意义吗？任何例子？
我的Eclipse RCP Maven项目使用Java 8构建得很好,但是在Java 9中失败了:
[INFO] Scanning for projects...
[WARNING] Could not start bundle org.eclipse.equinox.registry
org.osgi.framework.BundleException: Could not resolve module: org.eclipse.equinox.registry [42]
Unresolved requirement: Require-Bundle: org.eclipse.equinox.common; bundle-version="[3.7.0,4.0.0)"
-> Bundle-SymbolicName: org.eclipse.equinox.common; bundle-version="3.8.0.v20160315-1450"; singleton:="true"
org.eclipse.equinox.common [18]
Unresolved requirement: Require-Capability: osgi.ee; filter:="(&(osgi.ee=JavaSE)(version=1.7))"
Unresolved requirement: Import-Package: org.eclipse.core.runtime.jobs; resolution:="optional"
-> Export-Package: org.eclipse.core.runtime.jobs; bundle-symbolic-name="org.eclipse.core.jobs"; bundle-version="3.8.0.v20160209-0147"; version="0.0.0"
org.eclipse.core.jobs [6]
Unresolved requirement: Require-Bundle: org.eclipse.equinox.common; bundle-version="[3.8.0,4.0.0)"
-> Bundle-SymbolicName: org.eclipse.equinox.common; bundle-version="3.8.0.v20160315-1450"; singleton:="true"
at org.eclipse.osgi.container.Module.start(Module.java:434)
at org.eclipse.osgi.internal.framework.EquinoxBundle.start(EquinoxBundle.java:392)
at org.eclipse.sisu.equinox.embedder.internal.DefaultEquinoxEmbedder.tryActivateBundle(DefaultEquinoxEmbedder.java:215)
at org.eclipse.sisu.equinox.embedder.internal.DefaultEquinoxEmbedder.activateBundlesInWorkingOrder(DefaultEquinoxEmbedder.java:207)
at org.eclipse.sisu.equinox.embedder.internal.DefaultEquinoxEmbedder.doStart(DefaultEquinoxEmbedder.java:182)
at org.eclipse.sisu.equinox.embedder.internal.DefaultEquinoxEmbedder.start(DefaultEquinoxEmbedder.java:67)
at org.eclipse.sisu.equinox.embedder.internal.DefaultEquinoxEmbedder.checkStarted(DefaultEquinoxEmbedder.java:310)
at org.eclipse.sisu.equinox.embedder.internal.DefaultEquinoxEmbedder.getService(DefaultEquinoxEmbedder.java:286)
at org.eclipse.sisu.equinox.embedder.internal.DefaultEquinoxEmbedder.getService(DefaultEquinoxEmbedder.java:280) …是否有一种优雅的方式将其转换Map<P, Optional<Q>>为稀疏Map<P, Q>？
这应该工作,但它有点meh:
Map<P,Optional<Q>> map = ...;
Map<P,Q> map2 = map.entrySet()
.stream().filter(e -> e.getValue().isPresent())
.collect(Collectors.toMap(e -> e.getKey(), e->e.getValue().get()));
我有以下10000000x2矩阵:
0 0
1 1
2 2
.. ..
10000000 10000000
现在我想将这个矩阵保存到int[][]数组:
import com.google.common.base.Stopwatch;
static void memory(int size) throws Exception {
System.out.println("Memory");
Stopwatch s = Stopwatch.createStarted();
int[][] l = new int[size][2];
for (int i = 0; i < size; i++) {
l[i][0] = i;
l[i][1] = i;
}
System.out.println("Keeping " + size + " rows in-memory: " + s.stop());
}
public static void main(String[] args) throws Exception {
int size = 10000000;
memory(size);
memory(size);
memory(size);
memory(size);
memory(size);
} …match (str) {
case "String1" => ???
case "String2" => ???
}
这是区分大小写的匹配.如何编写不区分大小写的匹配？我知道我可以为每个分支调用toLowerCase,但我想要更优雅的解决方案.
我正在读JEP 317.它说Graal(一个新的基于实验Java的JIT编译器)将成为JDK 10的一部分,但后来它说已经在JDK 9中可用了.那么,JEP 317的重点是什么呢？Java 9是否包含Graal？
考虑以下几乎可编译的Java 8代码:
public static void main(String[] args) {
LinkedList<User> users = null;
users.add(new User(1, "User1"));
users.add(new User(2, "User2"));
users.add(new User(3, "User3"));
User user = users.stream().filter((user) -> user.getId() == 1).findAny().get();
}
static class User {
int id;
String username;
public User() {
}
public User(int id, String username) {
this.id = id;
this.username = username;
}
public void setUsername(String username) {
this.username = username;
}
public void setId(int id) {
this.id = id;
}
public String getUsername() {
return …鉴于:
Integer[] arr1 = {1, 5, 9, 17};
Integer[] arr2 = {1, 2, 3, 6, 7, 12, 15};
Observable<Integer> o1 = Observable.from(arr1);
Observable<Integer> o2 = Observable.from(arr2);
如何获得包含的Observable 1, 1, 2, 3, 5, 6, 7, 9, 12, 15, 17？