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;?????????}?????}?}??#?輸出

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?//?將常量1推入操作數(shù)棧頂??istore_0?//?彈出棧頂元素(1)，保存到局部變量表slot[0]，此時slot[0]=1。這兩條指令對應源碼：x?=?1;??iload_0?//?將局部變量表slot[0]的值推入操作數(shù)棧頂，也就是說把上面x的值推入棧頂??istore_1?//?彈出棧頂元素(1)，保存到局部變量表slot[1]，此時slot[1]=1。其實，此時就已經(jīng)把要return的值準備好了??iconst_3?//?將常量3推入操作數(shù)棧頂，這一條指令開始，其實是開始執(zhí)行finally中的代碼了??istore_0?//?彈出棧頂元素(3)，保存到局部變量表slot[0]，此時slot[0]=3。這兩個指令對應源碼：x?=?3;這里要注意的是，雖然都是更新了x的值，但是finally中的x和try中x的賦值，保存在了不同的局部變量表中?iload_1?//?將局部變量表slot[1]的值推入操作數(shù)棧頂，此時棧頂元素的值為1，是第3行指令保存的值??ireturn?//?將操作數(shù)棧頂?shù)闹捣祷亟o調(diào)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;?????}?}??#?輸出

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

iconst_1?//?將常量1推入操作數(shù)棧頂?istore_0?//?彈出棧頂元素（1），保存到局部變量表slot[0]，此時slot[0]=1。這兩條指令對應源碼：x?=?1;?iinc??????????0,?1?//?對局部變量表slot[0]進行自增（+1）操作，此時slot[0]=2，對應源碼：++x;所以，可以看出return后面的表達式先執(zhí)行?iload_0?//?將局部變量表slot[0]的值推入操作數(shù)棧頂，也就是說把上面x的值（2）推入棧頂?istore_1?//?彈出棧頂元素（2），保存到局部變量表slot[1]，此時slot[1]=2。其實，此時就已經(jīng)把要return的值準備好了?iconst_3?//?將常量3推入操作數(shù)棧頂，這一條指令開始，其實是開始執(zhí)行finally中的代碼了?istore_0?//?彈出棧頂元素（3），保存到局部變量表slot[0]，此時slot[0]=3。這兩個指令對應源碼：x?=?3;這里要注意的是，雖然都是更新了x的值，但是finally中的x和try中x的賦值，保存在了不同的局部變量表中?iload_1?//?將局部變量表slot[1]的值推入操作數(shù)棧頂，此時棧頂元素的值為2，是第6行指令保存的值，也就是經(jīng)過++x之后的值?ireturn?//?將操作數(shù)棧頂?shù)闹捣祷亟o調(diào)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;?????}?}?#?輸出

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?//?將常量1推入操作數(shù)棧頂?istore_0?//?彈出棧頂元素（1），保存到局部變量表slot[0]，此時slot[0]=1。這兩條指令對應源碼：x?=?1;?iinc??????????0,?1?//?對局部變量表slot[0]進行自增（+1）操作，此時slot[0]=2，對應源碼：++x;所以，可以看出return后面的表達式先執(zhí)行?iload_0??//?將局部變量表slot[0]的值推入操作數(shù)棧頂，也就是說把上面x的值（2）推入棧頂?istore_1?//?彈出棧頂元素（2），保存到局部變量表slot[1]，此時slot[1]=2。?iconst_3?//?將常量3推入操作數(shù)棧頂，這一條指令開始，其實是開始執(zhí)行finally中的代碼了?istore_0?//?彈出棧頂元素（3），保存到變量表slot[0]，此時slot[0]=3。這兩個指令對應源碼：x?=?3?iload_0??//?將局部變量表slot[0]的值（3）推入操作數(shù)棧，這是跟之前不一樣的地方，ireturn返回的值選擇的局部變量表不一樣?ireturn

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