java - int 的基本算术运算 - Java

Question

我最近注意到 Java 关于 Java 中基本算术运算的特性。用下面的代码

byte a = 3;
byte b = 4;
byte c = a * b;

我收到“类型不匹配”编译错误...

Java中的基本算术运算(+、-、*、/)只对primitive进行int 和高阶数据类型(long、double 等)，而 byte 和short 首先转换为 int 然后计算？

Answer 1

对 byte、char 和 short 的操作被扩展为 int，除非编译器可以确定该值是在范围内。

final byte a = 3, b = 4;
byte c = a * b; // compiles

final byte a = 3, b = 40;
byte c = a * b; // compiles

final int a = 3, b = 4;
byte c = a * b; // compiles !!

但是

byte a = 3, b = 4;
byte c = a * b; // doesn't compile as the result of this will be `int` at runtime.

final byte a = 30, b = 40;
byte c = a * b; // doesn't compile as the value is too large, will be an `int`

顺便说一句，即使它导致溢出，也会编译。 :]

final int a = 300000, b = 400000;
int c = a * b; // compiles but overflows, is not made a `long`