本帖最后由 jojo 于 2015-1-10 16:30 编辑
据说能够正确分析出下面这道程序的输出就证明你的Java异常完全没有任何问题了。本以为自己学得还可以,结果瞬间被秒杀了,有兴趣的童鞋可以试试。答案我会附在最后面。
- 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();
- }
- }
- }
复制代码
答案:
i=2
i=1
testEx2, catch exception
testEx2, finally; return value=false
testEx1, finally; return value=false
testEx, finally; return value=false
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