当使用“加号等于"时，Java为什么要执行从双精度到整数的隐式类型转换?操作员? [英] Why does Java perform implicit type conversion from double to integer when using the "plus equals" operator?

问题描述

可能重复:
因行为而变化，可能会导致精度损失

代码示例A

 public class Test {                                                         
     public static void main(String[] args) {
         int i = 0;
         i = i + 1.5;
     }
 }

代码示例B

 public class Test {                                                         
     public static void main(String[] args) {
         int i = 0;
         i += 1.5;
     }
 }

毫不奇怪，编译A会产生以下错误.令人惊讶的是，编译B不会产生任何错误，并且看起来就像我在double值1.5之前插入了显式强制转换为integer一样.为什么在世界上会发生这种情况?这违背了我以为我知道的一切！

Unsurprisingly, compiling A produces the error below. Surprisingly, compiling B produces no error and it appears to behave as if I inserted an explicit cast to integer before the double value 1.5. Why in the world does this happen? This goes against everything I thought I knew!

Test.java:6: possible

 loss of precision

    found   : double
    required: int
            i = i + 1.5;
                  ^
    1 error

推荐答案

它按设计工作.复合运算符向该运算符添加隐式强制转换.否则，您必须使用显式强制转换.

It is working as designed. The compound operators add an implicit cast to the operation. Otherwise you have to use an explicit cast.

更多信息?

http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.26.2

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