控制台上的其他消息未按正确顺序打印堆栈跟踪

Question

为什么 会不会发生为以下 Java 程序打印的 堆栈跟踪 未按正确顺序显示在控制台屏幕上？它与屏幕上的其他消息混在一起。

是否有任何并行性导致它？

Java 程序:[=3​​8=]

package evm;

public class Client {

    public static void main(String[] args) {
        EVM evm = new EVM();
        
        try {
            evm.setCandidates(90);   /**An Exception thrown here**/
        } catch (CandidatesOutOfLimitsException e) {
            e.printStackTrace();
            //System.out.print(e.getMessage());
        }
        
        try {
            evm.voteForCandidate(43);    /**An Exception thrown here**/
        } catch (BallotUnitOffException e1) {
            e1.printStackTrace();
            //System.out.print(e1.getMessage());
        }

        evm.pressBallotButton();
        
        System.out.println(evm);  //other messages
        evm.switchOn();
        System.out.println(evm);  //other messages
        
        try {
            evm.voteForCandidate(43);    /**An Exception thrown here**/
        } catch (BallotUnitOffException e) {
            e.printStackTrace();
            //System.out.print(e.getMessage());
        }
    }

}

我对抛出异常的行进行了评论。

运行 1:

evm.CandidatesOutOfLimitsException: Number of Candidates cannot exceed 64
    at evm.EVM.setCandidates(EVM.java:41)
    at evm.Client.main(Client.java:9)
evm.BallotUnitOffException: Ballot Unit is not On
    at evm.BallotUnit.pressCandidateButton(BallotUnit.java:38)
    at evm.EVM.voteForCandidate(EVM.java:59)
    at evm.Client.main(Client.java:16)
evm.BallotUnitOffException: Ballot Unit is not On
    at evm.BallotUnit.pressCandidateButton(BallotUnit.java:38)
    at evm.EVM.voteForCandidate(EVM.java:59)
    at evm.Client.main(Client.java:28)

Control Unit State: evm.Off@42a57993
On Lamp: evm.Off@15db9742Ballot Unit: Ready Lamp: evm.Off@6d06d69c
Slide Switch:evm.SlideSwitchOne@7852e922
Ballot Unit: Ready Lamp: evm.Off@4e25154f
Slide Switch:evm.SlideSwitchTwo@70dea4e
Ballot Unit: Ready Lamp: evm.Off@5c647e05
Slide Switch:evm.SlideSwitchThree@33909752
Ballot Unit: Ready Lamp: evm.Off@55f96302
Slide Switch:evm.SlideSwitchFour@3d4eac69


Control Unit State: evm.On@28d93b30
On Lamp: evm.On@75b84c92Ballot Unit: Ready Lamp: evm.On@6bc7c054
Slide Switch:evm.SlideSwitchOne@7852e922
Ballot Unit: Ready Lamp: evm.On@232204a1
Slide Switch:evm.SlideSwitchTwo@70dea4e
Ballot Unit: Ready Lamp: evm.On@4aa298b7
Slide Switch:evm.SlideSwitchThree@33909752
Ballot Unit: Ready Lamp: evm.On@7d4991ad
Slide Switch:evm.SlideSwitchFour@3d4eac69

运行 2:

evm.CandidatesOutOfLimitsException: Number of Candidates cannot exceed 64
    at evm.EVM.setCandidates(EVM.java:41)
    at evm.Client.main(Client.java:9)
evm.BallotUnitOffException: Ballot Unit is not On
    at evm.BallotUnit.pressCandidateButton(BallotUnit.java:38)
    at evm.EVM.voteForCandidate(EVM.java:59)
    at evm.Client.main(Client.java:16)

Control Unit State: evm.Off@42a57993
On Lamp: evm.Off@15db9742Ballot Unit: Ready Lamp: evm.Off@6d06d69c
Slide Switch:evm.SlideSwitchOne@7852e922
Ballot Unit: Ready Lamp: evm.Off@4e25154f
Slide Switch:evm.SlideSwitchTwo@70dea4e
Ballot Unit: Ready Lamp: evm.Off@5c647e05
Slide Switch:evm.SlideSwitchThree@33909752
Ballot Unit: Ready Lamp: evm.Off@55f96302
Slide Switch:evm.SlideSwitchFour@3d4eac69


Control Unit State: evm.On@28d93b30
On Lamp: evm.On@75b84c92Ballot Unit: Ready Lamp: evm.On@6bc7c054
Slide Switch:evm.SlideSwitchOne@7852e922
Ballot Unit: Ready Lamp: evm.On@232204a1
Slide Switch:evm.SlideSwitchTwo@70dea4e
Ballot Unit: Ready Lamp: evm.On@4aa298b7
Slide Switch:evm.SlideSwitchThree@33909752
Ballot Unit: Ready Lamp: evm.On@7d4991ad
Slide Switch:evm.SlideSwitchFour@3d4eac69

evm.BallotUnitOffException: Ballot Unit is not On
    at evm.BallotUnit.pressCandidateButton(BallotUnit.java:38)
    at evm.EVM.voteForCandidate(EVM.java:59)
    at evm.Client.main(Client.java:28)

同样，我每次 运行 都会得到一些其他模式。任何人解释这种行为。

我正在使用

Eclipse Java EE IDE for Web Developers.

Version: Luna Release (4.4.0)

Build id: 20140612-0600

对于冗长的问题，我深表歉意

Answer 1

与 Eclipse 相比，这与标准输出和错误流的实现有关。

System.out 和 System.err 都是 PrintStream 对象，它们可以传递缓冲流，不会自动将数据刷新到它们的目的地。由于它们是两个内部使用缓冲的不同流，您写入它们的数据可能会交错，因为流可能决定在不同时间刷新（缓冲区内容可能在不同时间填充）。

在打印前两个堆栈跟踪后调用 System.err.flush()，然后在打印到标准输出流后调用 System.out.flush() 应该会得到一致的结果。