考虑到这段代码,我能绝对确定finally块总是执行的吗,不管something()是什么?
try {
something();
return success;
}
catch (Exception e) {
return failure;
}
finally {
System.out.println("I don't know if this will get printed out");
}
考虑到这段代码,我能绝对确定finally块总是执行的吗,不管something()是什么?
try {
something();
return success;
}
catch (Exception e) {
return failure;
}
finally {
System.out.println("I don't know if this will get printed out");
}
当前回答
尝试这段代码,您将了解finally块中的代码是在return语句之后执行的。
public class TestTryCatchFinally {
static int x = 0;
public static void main(String[] args){
System.out.println(f1() );
System.out.println(f2() );
}
public static int f1(){
try{
x = 1;
return x;
}finally{
x = 2;
}
}
public static int f2(){
return x;
}
}
其他回答
除其他响应外,重要的是要指出,“finally”有权通过try重写任何异常/返回值。。捕捉块。例如,以下代码返回12:
public static int getMonthsInYear() {
try {
return 10;
}
finally {
return 12;
}
}
类似地,以下方法不会引发异常:
public static int getMonthsInYear() {
try {
throw new RuntimeException();
}
finally {
return 12;
}
}
虽然以下方法确实抛出了它:
public static int getMonthsInYear() {
try {
return 12;
}
finally {
throw new RuntimeException();
}
}
是的,会的。唯一不会发生的情况是JVM退出或崩溃
如果在嵌套的finally块中引发异常,finally也可以提前退出。编译器会警告你finally块没有正常完成,或者给出一个错误,说明你有无法访问的代码。仅当抛出不在条件语句后面或循环内部时,才会显示不可访问代码的错误。
try{
}finally{
try{
}finally{
//if(someCondition) --> no error because of unreachable code
throw new RunTimeException();
}
int a = 5;//unreachable code
}
下面是凯文的回答。重要的是要知道,要返回的表达式在finally之前求值,即使在finally之后返回。
public static void main(String[] args) {
System.out.println(Test.test());
}
public static int printX() {
System.out.println("X");
return 0;
}
public static int test() {
try {
return printX();
}
finally {
System.out.println("finally trumps return... sort of");
return 42;
}
}
输出:
X
finally trumps return... sort of
42
答案很简单:是。
输入:
try{
int divideByZeroException = 5 / 0;
} catch (Exception e){
System.out.println("catch");
return; // also tried with break; in switch-case, got same output
} finally {
System.out.println("finally");
}
输出:
catch
finally