Java:复合赋值和基于表达式的类型提升

Question

在下面的三个按位左移代码片段中，Java 对示例 #2 和 #3 的处理方式不同，这很有趣。在最后一个示例(#3)中，为什么 Java 决定不将复合赋值语句升级为 int？

答案是否与 Java“内联”做事有关。非常感谢任何评论。

byte b = -128;

// Eg #1.  Expression is promoted to an int, and its expected value for an int is -256.
System.out.println(b << 1);

b = -128;
// Eg #2.  Must use a cast, otherwise a compilation error will occur.  
// Value is 0, as to be expected for a byte.
System.out.println(b = (byte)(b << 1));

b = -128;
// Eg #3.  Not only is no cast required, but the statement isn't "upgraded" to an int.
// Its value is 0, as to be expected for a byte.
System.out.println(b <<= 1);

Answer 1

复合赋值运算符，例如 +=和 -=和 <<=等在它们的操作中有一个隐式类型转换。

换句话说。

byte x = 1;
x <<= 4;

等于:

byte x = 1;
x = (byte)(x << 4);

编译时。

左移操作仍然适本地提升变量(在 byte 的情况下为 int )但复合赋值运算符为您转换它。