What is the execution order of return and finally statements in Java?

Source code:

public class ReturnFinallyDemo {     public static void main(String[] args) {         System.out.println(case1());     }      public static int case1() {         int x;         try {             x = 1;             return x;         } finally {             x = 3;         }     } }  # 输出

The output of the above code can simply conclude: return is executed before finally. Let’s take a look at what happens at the bytecode level. The following intercepts part of the bytecode of the case1 method, and compares it with the source code to annotate the meaning of each instruction:

iconst_1 // 将常量1推入操作数栈顶  istore_0 // 弹出栈顶元素(1)，保存到局部变量表slot[0]，此时slot[0]=1。这两条指令对应源码：x = 1;  iload_0 // 将局部变量表slot[0]的值推入操作数栈顶，也就是说把上面x的值推入栈顶  istore_1 // 弹出栈顶元素(1)，保存到局部变量表slot[1]，此时slot[1]=1。其实，此时就已经把要return的值准备好了  iconst_3 // 将常量3推入操作数栈顶，这一条指令开始，其实是开始执行finally中的代码了  istore_0 // 弹出栈顶元素(3)，保存到局部变量表slot[0]，此时slot[0]=3。这两个指令对应源码：x = 3;这里要注意的是，虽然都是更新了x的值，但是finally中的x和try中x的赋值，保存在了不同的局部变量表中 iload_1 // 将局部变量表slot[1]的值推入操作数栈顶，此时栈顶元素的值为1，是第3行指令保存的值  ireturn // 将操作数栈顶的值返回给调用方

From the bytecode, it seems that the finally code is executed first, because ireturn The instruction is indeed executed at the end, so the value returned does not depend on who executes it first, but on when the top element of the operand stack returned by the ireturn instruction is saved. In the above code environment, it is the version assigned to x in the try code block, that is, the version of x saved immediately after the return statement.

Let’s look at a slightly more complicated scenario:

public static int case2() {     int x;     try {         x = 1;         return ++x;     } finally {         x = 3;     } }  # 输出

With the above analysis, this is easy to understand. Let’s take a look at the bytecode:

iconst_1 // 将常量1推入操作数栈顶 istore_0 // 弹出栈顶元素（1），保存到局部变量表slot[0]，此时slot[0]=1。这两条指令对应源码：x = 1; iinc          0, 1 // 对局部变量表slot[0]进行自增（+1）操作，此时slot[0]=2，对应源码：++x;所以，可以看出return后面的表达式先执行 iload_0 // 将局部变量表slot[0]的值推入操作数栈顶，也就是说把上面x的值（2）推入栈顶 istore_1 // 弹出栈顶元素（2），保存到局部变量表slot[1]，此时slot[1]=2。其实，此时就已经把要return的值准备好了 iconst_3 // 将常量3推入操作数栈顶，这一条指令开始，其实是开始执行finally中的代码了 istore_0 // 弹出栈顶元素（3），保存到局部变量表slot[0]，此时slot[0]=3。这两个指令对应源码：x = 3;这里要注意的是，虽然都是更新了x的值，但是finally中的x和try中x的赋值，保存在了不同的局部变量表中 iload_1 // 将局部变量表slot[1]的值推入操作数栈顶，此时栈顶元素的值为2，是第6行指令保存的值，也就是经过++x之后的值 ireturn // 将操作数栈顶的值返回给调用方

As can be seen from the above code, the instruction after return is executed first, then saved to the local variable table, then the statement in finally is executed, and finally the return instruction itself is executed.

To summarize, the return instruction is executed last. If there is an expression after return, the statement in finally will be executed after the expression is executed, and the return instruction will be executed last. So which one is executed first, finally or return: If there is an expression or method call after the return instruction, it will be executed first, then finally, and finally the return instruction. Just like the result of the above program demonstration, the final return result cannot be judged just from the assignment of x. From the instruction level, the two assignments to x are stored in different locations in the local variable table.

Finally, let’s look at a scenario that is not usually written like this:

public static int case3() {     int x;     try {         x = 1;         return ++x;     } finally {         x = 3;         return x;     } } # 输出

This is an example of finally returning results. It is not usually recommended to write like this. We also look at it from the perspective of bytecode Let’s analyze:

iconst_1 // 将常量1推入操作数栈顶 istore_0 // 弹出栈顶元素（1），保存到局部变量表slot[0]，此时slot[0]=1。这两条指令对应源码：x = 1; iinc          0, 1 // 对局部变量表slot[0]进行自增（+1）操作，此时slot[0]=2，对应源码：++x;所以，可以看出return后面的表达式先执行 iload_0  // 将局部变量表slot[0]的值推入操作数栈顶，也就是说把上面x的值（2）推入栈顶 istore_1 // 弹出栈顶元素（2），保存到局部变量表slot[1]，此时slot[1]=2。 iconst_3 // 将常量3推入操作数栈顶，这一条指令开始，其实是开始执行finally中的代码了 istore_0 // 弹出栈顶元素（3），保存到变量表slot[0]，此时slot[0]=3。这两个指令对应源码：x = 3 iload_0  // 将局部变量表slot[0]的值（3）推入操作数栈，这是跟之前不一样的地方，ireturn返回的值选择的局部变量表不一样 ireturn

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