在 Java 中，后自增运算符在 return 语句中如何起作用？

发布于 2024-07-14 18:32:24 字数 263 浏览 7 评论 0原文

给定以下代码， ixAdd 是否会执行您所期望的操作，即在递增之前返回 ix 的值，但在离开函数之前递增类成员？

class myCounter {
    private int _ix = 1;

    public int ixAdd()
    {
        return _ix++;
    }
}

当程序离开函数的堆栈帧（或 Java 中的任何内容）时，我不太确定后/前增量的常见规则是否也适用于 return 语句。

原文

Given the following code, will ixAdd do what you'd expect, i. e. return the value of ix before the increment, but increment the class member before leaving the function?

class myCounter {
    private int _ix = 1;

    public int ixAdd()
    {
        return _ix++;
    }
}

I wasn't quite sure if the usual rules for post / pre increment would also apply in return statements, when the program leaves the stack frame (or whatever it is in Java) of the function.

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ヤ经典坏疍 2024-07-21 18:32:24

关键部分是在表达式求值后立即发生后递增/递减。它不仅发生在返回发生之前 - 它发生在计算任何后续表达式之前。例如，假设您写道：

class myCounter {
    private int _ix = 1;

    public int ixAdd()
    {
        return _ix++ + giveMeZero();
    }

    public int giveMeZero()
    {
        System.out.println(_ix);
        return 0;
    }
}

这也会打印出递增的结果，因为递增发生在调用 giveMeZero() 之前。

The key part is that a post increment/decrement happens immediately after the expression is evaluated. Not only does it happen before the return occurs - it happens before any later expressions are evaluated. For instance, suppose you wrote:

class myCounter {
    private int _ix = 1;

    public int ixAdd()
    {
        return _ix++ + giveMeZero();
    }

    public int giveMeZero()
    {
        System.out.println(_ix);
        return 0;
    }
}

That would print out the incremented result as well, because the increment happens before giveMeZero() is called.

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爱你不解释 2024-07-21 18:32:24

好吧，让我们看一下字节码（使用javap -c自己看看）：

Compiled from "PostIncTest.java"
class myCounter extends java.lang.Object{
myCounter();
  Code:
   0:   aload_0
   1:   invokespecial   #1; //Method java/lang/Object."<init>":()V
   4:   aload_0
   5:   iconst_1
   6:   putfield    #2; //Field _ix:I
   9:   return

public int ixAdd();
  Code:
   0:   aload_0
   1:   dup
   2:   getfield    #2; //Field _ix:I
   5:   dup_x1
   6:   iconst_1
   7:   iadd
   8:   putfield    #2; //Field _ix:I
   11:  ireturn

}

如您所见，ixAdd()中的指令6和7处理返回前的增量。因此，正如我们所期望的，递减运算符在 return 语句中使用时确实有效果。但请注意，在这两个指令出现之前有一个 dup 指令；增加的值（当然）不会反映在返回值中。

Well, let's look at the bytecode (use javap -c <classname> to see it yourself):

Compiled from "PostIncTest.java"
class myCounter extends java.lang.Object{
myCounter();
  Code:
   0:   aload_0
   1:   invokespecial   #1; //Method java/lang/Object."<init>":()V
   4:   aload_0
   5:   iconst_1
   6:   putfield    #2; //Field _ix:I
   9:   return

public int ixAdd();
  Code:
   0:   aload_0
   1:   dup
   2:   getfield    #2; //Field _ix:I
   5:   dup_x1
   6:   iconst_1
   7:   iadd
   8:   putfield    #2; //Field _ix:I
   11:  ireturn

}

As you can see, instructions 6 and 7 in ixAdd() handle the increment before the return. Therefore, as we would expect, the decrement operator does indeed have an effect when used in a return statement. Notice, however, that there is a dup instruction before these two appear; the incremented value is (of course) not reflected in the return value.

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