为什么对空 Atomic class 调用方法不会产生异常?
Why does invoking a method to a null Atomic class does not produce exception?
import java.util.concurrent.*;
import java.util.concurrent.atomic.AtomicInteger;
public class _7_Synchronizing_Data_Access {
private AtomicInteger count;
private void incrementAndReport() {
System.out.print(count.incrementAndGet() + "here"); //does not print
}
public static void main(String[] args) {
ExecutorService service = null;
try {
service = Executors.newFixedThreadPool(20);
_7_Synchronizing_Data_Access manager = new _7_Synchronizing_Data_Access();
for (int i = 0; i < 10; i++)
service.submit(() -> manager.incrementAndReport());
} finally {
if (service != null)
service.shutdown();
}
}
}
运行 这个程序没有输出。甚至没有 NullPointerException。如您所见,我没有实例化 count
。我希望它会引发错误。这是为什么?
抛出并捕获了 NullPointerException
异常,但为了查看它们,您需要检查 service.submit
调用返回的 Future
个实例:
按如下方式更改循环:
for (int i = 0; i < 10; i++) {
Future f = service.submit(() -> manager.incrementAndReport());
try {
System.out.println (f.get ());
}
catch (ExecutionException exex) {
System.out.println (exex);
}
catch (InterruptedException intEx) {
System.out.println (intEx);
}
}
将输出:
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
如果用 try-catch:
包围 System.out.println(count.incrementAndGet() + "here");
语句,您还会看到会抛出异常
private void incrementAndReport() {
try {
System.out.println(count.incrementAndGet() + "here"); //does not print
}
catch (Exception exc) {
System.out.println (exc);
}
}
import java.util.concurrent.*;
import java.util.concurrent.atomic.AtomicInteger;
public class _7_Synchronizing_Data_Access {
private AtomicInteger count;
private void incrementAndReport() {
System.out.print(count.incrementAndGet() + "here"); //does not print
}
public static void main(String[] args) {
ExecutorService service = null;
try {
service = Executors.newFixedThreadPool(20);
_7_Synchronizing_Data_Access manager = new _7_Synchronizing_Data_Access();
for (int i = 0; i < 10; i++)
service.submit(() -> manager.incrementAndReport());
} finally {
if (service != null)
service.shutdown();
}
}
}
运行 这个程序没有输出。甚至没有 NullPointerException。如您所见,我没有实例化 count
。我希望它会引发错误。这是为什么?
抛出并捕获了 NullPointerException
异常,但为了查看它们,您需要检查 service.submit
调用返回的 Future
个实例:
按如下方式更改循环:
for (int i = 0; i < 10; i++) {
Future f = service.submit(() -> manager.incrementAndReport());
try {
System.out.println (f.get ());
}
catch (ExecutionException exex) {
System.out.println (exex);
}
catch (InterruptedException intEx) {
System.out.println (intEx);
}
}
将输出:
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
java.util.concurrent.ExecutionException: java.lang.NullPointerException
如果用 try-catch:
包围System.out.println(count.incrementAndGet() + "here");
语句,您还会看到会抛出异常
private void incrementAndReport() {
try {
System.out.println(count.incrementAndGet() + "here"); //does not print
}
catch (Exception exc) {
System.out.println (exc);
}
}