为什么 Java 中的增广赋值运算符会给出不同的结果? [英] Why does the augmented assignement operator in Java give a different result?

问题描述

考虑以下代码:

class 转换器 {公共静态无效主(字符串参数[]){字节数 = 1;数量 = 数量 * 2.5；System.out.println("结果为:" + num);}}

编译器抛出以下错误:

错误:不兼容的类型:第 1 行从双精度到字节的可能有损转换

如果我更改了 main() 方法的第二行并使用了 *= 速记运算符:

类转换器{公共静态无效主(字符串参数[]){字节数 = 1;数量 *= 2.5;System.out.println("结果为:" + num);}}

代码编译并成功运行，输出:

结果是:2

为什么 *= 速记运算符与完整表达式 num = num * 2.5 的行为不同?

解决方案

来自 JLS 复合赋值运算符 文档，你可以看下面的例子:

<块引用>

例如下面的代码是正确的:

short x = 3;x += 4.6；

<块引用>

并导致 x 的值为 7，因为它等价于:

short x = 3;x = (短)(x + 4.6);

默认情况下它只是自动转换结果.

PS:在这个其他答案 我试图解释您需要使用 JLS 进行如下操作的原因

short x = 3;x = x + 1;//不会编译

它是全新的，所以我乐于接受建议.

Consider the following piece of code:

class Converter {
    public static void main(String args[]) {
        byte num = 1;
        num = num * 2.5;
        System.out.println("Result is: " + num);
    }
}

The compiler throws the following error:

error: incompatible types: possible lossy conversion from double to the byte at line 1

If I change the second line of the main() method and I use the *= shorthand operator:

class Converter {
    public static void main(String args[]) {
        byte num = 1;
        num *= 2.5;
        System.out.println("Result is: " + num);
    }
}

The code compiles and runs successfully with the output:

Result is: 2

Why the *= shorthand operator is behaving differently from the full expression num = num * 2.5?

解决方案

From the JLS compound assigment operator documentation, you can see the following example :

For example, the following code is correct:

short x = 3;
x += 4.6;

and results in x having the value 7 because it is equivalent to:

short x = 3;
x = (short)(x + 4.6);

It simply auto cast the result by default.

PS : In this other answer I tried to explain the reason you need to cast an operation like the following using the JLS

short x = 3;
x = x + 1; //Won't compile

It is realativly new so I am open to suggestion there.

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