JAVA異常處理之finally中最好不要使用return

finally 語句塊中, 最好不要使用return, 否則會造成已下后果;

1, 如果catch塊中捕獲了異常, 並且在catch塊中將該異常throw給上級調用者進行處理, 但finally中return了, 那么catch塊中的throw就失效了, 上級方法調用者是捕獲不到異常的. 見demo如下:

public class TestException {
    public TestException() {
    }

    boolean testEx() throws Exception {
        boolean ret = true;
        try {
            ret = testEx1();
        } catch (Exception e) {
            System.out.println("testEx, catch exception");
            ret = false;
            throw e;
        } finally {
            System.out.println("testEx, finally; return value=" + ret);
            return ret;
        }
    }

    boolean testEx1() throws Exception {
        boolean ret = true;
        try {
            ret = testEx2();
            if (!ret) {
                return false;
            }
            System.out.println("testEx1, at the end of try");
            return ret;
        } catch (Exception e) {
            System.out.println("testEx1, catch exception");
            ret = false;
            throw e;
        } finally {
            System.out.println("testEx1, finally; return value=" + ret);
            return ret;
        }
    }

    boolean testEx2() throws Exception {
        boolean ret = true;
        try {
            int b = 12;
            int c;
            for (int i = 2; i >= -2; i--) {
                c = b / i;
                System.out.println("i=" + i);
            }
            return true;
        } catch (Exception e) {
            System.out.println("testEx2, catch exception");
            ret = false;
            throw e;
        } finally {
            System.out.println("testEx2, finally; return value=" + ret);
            return ret;
        }
    }

    public static void main(String[] args) {
        TestException testException1 = new TestException();
        try {
            testException1.testEx();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

該方法的最終執行結果如下:

i=2
i=1
testEx2, catch exception
testEx2, finally; return value=false
testEx1, finally; return value=false
testEx, finally; return value=false

 

盡管testEx2中的catch拋出異常, 但因為finally中return了, 那么testEx1中是捕獲不到異常的, 所以testEx1中的catch塊是不會執行的.

 

2, 如果在finally里的return之前執行了其它return , 那么最終的返回值是finally中的return:

 

public class TestException3 {
    public static int x = 3;
    int testEx() throws Exception {
        try {
            x = 4;
            return x;
        } catch (Exception e) {

        } finally {
            x = 5;
            return x;
        }
    }
    public static void main(String[] args) {
        TestException3 testException3 = new TestException3();
        try {
            int a = testException3.testEx();
            System.out.println(a);
            System.out.println(x);
        } catch (Exception e) {
        }
    }
}

輸出結果為5, 5

 

如果finally中的return被注釋掉, 那么輸出結果是4, 5