关于java中关闭流疑问-PHP中文网问答

下面这段代码的作用是压缩,会用到ZipOutputStream

    ZipOutputStream zipOutputStream = null;
    try {
      zipOutputStream = new ZipOutputStream(new FileOutputStream(zipPath));
    } catch (FileNotFoundException e) {
        throw new IllegalArgumentException("zipPath error " + "", e);
    } finally {
        if (null != zipOutputStream) {
            try {
                zipOutputStream.close();
            } catch (IOException e) {
                logger.error("", e);
            }
        }
    }

错误放大版1:

    ZipOutputStream zipOutputStream = null;
    try {
        FileOutputStream fileOutputStream =   new FileOutputStream(zipPath);
        throw new RuntimeException();
        zipOutputStream = new ZipOutputStream(fileOutputStream);
    } catch (FileNotFoundException e) {
        throw new IllegalArgumentException("zipPath error " + "", e);
    } finally {
        if (null != zipOutputStream) {
            try {
                zipOutputStream.close();
            } catch (IOException e) {
                logger.error("", e);
            }
        }
    }

错误放大版2:

    ZipOutputStream zipOutputStream = null;
    try {
      zipOutputStream = new ZipOutputStream(new FileOutputStream(zipPath),null);
    } catch (FileNotFoundException e) {
        throw new IllegalArgumentException("zipPath error " + "", e);
    } finally {
        if (null != zipOutputStream) {
            try {
                zipOutputStream.close();
            } catch (IOException e) {
                logger.error("", e);
            }
        }
    }

这里我有一个疑问,在new FileOutputStream(zipPath)的过程中如果已经打开了这个文件,但是在new ZipOutputStream过程中出错了,这个时候zipOutputStream并没有创建成功,所以在finally中不会调用close方法,造成的结果就是这个文件不会被关闭。
所以这种关闭方式是不是会有漏洞,还是我的理由有问题?

/** * @param args * @throws Exception */ public static void main(String[] args) throws Exception { Test test = new Test(); test.test1(); System.out.println(123); } public void test1() { try { ArrayList<Double> al = new ArrayList<Double>(); for (int i = 0; i < 1000000; i++) { for (int j = 0; j < 1000000; j++) { al.add(Math.random()); } } } catch (Throwable e) { System.out.println("catch"); } finally { System.out.println("finally"); } }

finally Exception in thread "main" java.lang.OutOfMemoryError: Java heap space at java.lang.Double.valueOf(Unknown Source) at qmk.TestAnything.test1(TestAnything.java:50) at qmk.TestAnything.main(TestAnything.java:41)

全部回复(3)

迷茫2017-04-17 16:41:04 3楼

这个例子有漏洞，不过漏洞不在这里，而是 FileOutputStream。
如果 new ZipOutputStream()调用失败，这个构造方法会抛出运行时异常，而且不会关闭传递进来的 FileOutputStream 对象。
由于调用者也不持有该对象的引用，也无法释放资源，会造成打开的文件描述符没有关闭。
如果使用java8的话可以用 try with，或者在finally块里面释放FileOutputStream 对象持有的资源，这样是比较保险的做法。

赞 +0

添加回复