This article mainly introduces a brief analysis of the execution sequence of the try finally return statement in Java. Friends in need can refer to it
Problem Analysis
Will the finally statement block be executed?
Perhaps the first reaction of many people is that they will definitely implement it, but if they think about it carefully, if they will definitely implement it, they will not ask such SB.
Demo1
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public class Test { public static void main(String[] args) { System.out.println("return value of test(): " + test()); } public static int test() { int i = 1; // if (i == 1) { // return 0; // } System.out.println("the previous statement of try block"); i = i / 0; try { System.out.println("try block"); return i; } finally { System.out.println("finally block"); } } }The execution results of Demo1 are as follows:
the previous statement of try block Exception in thread "main" java.lang.ArithmeticException: / by zero at com.becoda.bkms.bus.basics.web.Test2.test(Test2.java:15) at com.becoda.bkms.bus.basics.web.Test2.main(Test2.java:5)In addition, if the comments in the above example are removed, the execution result is:
return value of test(): 0In the above two cases, the finally statement block is not executed. What problem does it explain? The finally statement block will be executed only when the try statement block corresponding to the finally statement block is executed. However, the above all return (return) or throw an exception before the try statement block, so the finally statement block corresponding to the try statement is not executed. So, even if the try statement block corresponding to finally is executed, will the finally statement block definitely be executed? But the execution results of the following example
Demo2
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public class Test { public static void main(String[] args) { System.out.println("return value of test(): " + test()); } public static int test() { int i = 1; try { System.out.println("try block"); System.exit(0); return i; } finally { System.out.println("finally block"); } } }
try block
Finally statement example explanationLet’s look at a simple example
Demo3
public class Test { public static void main(String[] args) { try { System.out.println("try block"); return; } finally { System.out.println("finally block"); } } }
try block finally block
Demo4
##public class Test { public static void main(String[] args) { System.out.println("reture value of test() : " + test()); } public static int test() { int i = 1; try { System.out.println("try block"); i = 1 / 0; return 1; } catch (Exception e) { System.out.println("exception block"); return 2; } finally { System.out.println("finally block"); } } }
The execution result of Demo4 is:
try block exception block finally block reture value of test() : 2
Demo4 illustrates that the finally statement block is executed before the return statement in the catch statement block.
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public class Test { public static void main(String[] args) { System.out.println("return value of getValue(): " + getValue()); } public static int getValue() { try { return 0; } finally { return 1; } } }Execution of Demo5 The result is:
return value of getValue(): 1
Demo6
public class Test { public static void main(String[] args) { System.out.println("return value of getValue(): " + getValue()); } public static int getValue() { int i = 1; try { return i; } finally { i++; } } }# The execution result of ##Demo6 is:
return value of getValue(): 1
Using the conclusion drawn from our above analysis: the finally statement block is executed before the return statement in try or catch. From this, we can easily understand that the execution result of Demo5 is 1. Because the return 1; statement in finally is executed before the return 0; statement in try, then after the return 1; statement in finally is executed, the control of the program is transferred to its caller main() function and returns The value is 1. So why is the return value of Demo6 not 2, but 1? According to the analysis logic of Demo5, the i++; statement in finally should be executed before return i; in try? The initial value of i is 1, then execute i++; then it is 2, and then execute return i; shouldn't it be 2? How did it become 1?
To explain this problem, you need to understand how the Java virtual machine compiles the finally statement block.
try{ return expression; }finally{ do some work; }
First We know that finally statements will definitely be executed, but what is their execution order? Their execution order is as follows:
1. Execution: expression, calculate the expression, and the result is saved on the top of the operand stack;
3、执行:finally语句块中的代码;
4、执行:将第2步复制到局部变量区的返回值又复制回操作数栈顶;
5、执行:return指令,返回操作数栈顶的值;
我们可以看到,在第一步执行完毕后,整个方法的返回值就已经确定了,由于还要执行finally代码块,因此程序会将返回值暂存在局部变量区,腾出操作数栈用来执行finally语句块中代码,等finally执行完毕,再将暂存的返回值又复制回操作数栈顶。所以无论finally语句块中执行了什么操作,都无法影响返回值,所以试图在finally语句块中修改返回值是徒劳的。因此,finally语句块设计出来的目的只是为了让方法执行一些重要的收尾工作,而不是用来计算返回值的。
这样就能解释Demo6的问题了
让我们再来看以下 3 个例子。
Demo7
public class Test { public static void main(String[] args) { System.out.println("return value of getValue(): " + getValue()); } @SuppressWarnings("finally") public static int getValue() { int i = 1; try { i = 4; } finally { i++; return i; } } }
Demo7的执行结果为:
return value of getValue(): 5
Demo8
public class Test { public static void main(String[] args) { System.out.println("return value of getValue(): " + getValue()); } public static int getValue() { int i = 1; try { i = 4; } finally { i++; } return i; } }
Demo8的执行结果为:
return value of getValue(): 5
Demo9
public class Test { public static void main(String[] args) { System.out.println(test()); } public static String test() { try { System.out.println("try block"); return test1(); } finally { System.out.println("finally block"); } } public static String test1() { System.out.println("return statement"); return "after return"; } }
Demo9的执行结果为:
try block return statement finally block after return
总结:
1、finally 语句块不一定会被执行
2、finally 语句块在 try 语句块中的 return 语句之前执行
3、finally 语句块在 catch 语句块中的 return 语句之前执行
4、finally 语句块中的 return 语句会覆盖 try 块中的 return 返回
5、试图在 finally 语句块中修改返回值不一定会被改变
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