如何在带有Try-catch的非返回方法中避免无限递归

Question

公共类示例｛

public static void main(String[] args) {
method();
}
public static void method()
{
try {
System.out.println("function");
throw new StaleElementReferenceException("thih sexception occured");
}
catch (StaleElementReferenceException e) {
method();
}
catch (Exception e) {
System.out.println("AssertFail");
}
}

}

如何在带有Try-catch的非返回方法中避免无限递归。。。例如下面的代码。。。当StaleElementException只发生一次时；函数，如果Stale元素第二次出现，我希望它转到Exception catch并打印Assert fail。。怎样

Answer 1

public class Sample {
public static void main(String[] args) {
method(false);
}
public static void method(boolean calledFromCatchBlock)
{
try {
System.out.println("function");
if(!calledFromCatchBlock) {
throw new StaleElementReferenceException("thih sexception occured");
} else {
throw new Exception();
}
} catch (StaleElementReferenceException e) {
method(true);
} catch (Exception e) {
System.out.println("AssertFail");
}
}
}

Answer 2

当您在method()外部抛出异常(例如boolean标志(时，您应该以某种方式存储状态，检查此状态并在下次抛出修改后的异常：

private static boolean alreadyThrown = false;
public static void method()
{
try {
System.out.println("function");
if (alreadyThrown) {
throw new RuntimeException("another exception occured");
} else {
alreadyThrown = true;
throw new StaleElementReferenceException("this exception occured");
}
}
catch (StaleElementReferenceException e) {
method();
}
catch (Exception e) {
System.out.println("AssertFail");
}
}

或者，您可以为method(int arg)提供一些参数，并以类似的方式检查其值：

public static void main(String[] args) {
method(1);
}
public static void method(int arg)
{
try {
System.out.println("function");
if (arg > 1) {
throw new RuntimeException("another exception occured");
} else {
throw new StaleElementReferenceException("this exception occured");
}
}
catch (StaleElementReferenceException e) {
method(arg + 1);
}
catch (Exception e) {
System.out.println("AssertFail");
}
}