Java隐式转换

使用以下代码：

Float a = 1.2;

有一个错误,因为它将十进制作为double值,double是一个比float更大的数据类型.

现在,它将整数作为默认的int类型.那么,为什么以下代码没有给出任何错误？

Byte b = 20;

解决方法

编译器非常聪明,可以确定20的位表示(int值)可以适合一个字节,而不会丢失数据.从 Java Language Specification §5.1.3：

A narrowing primitive conversion from double to float is governed by the IEEE 754 rounding rules (07001). This conversion can lose precision,but also lose range,resulting in a float zero from a nonzero double and a float infinity from a finite double. A double NaN is converted to a float NaN and a double infinity is converted to the same-signed float infinity.

A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits,where n is the number of bits used to represent type T. In addition to a possible loss of information about the magnitude of the numeric value,this may cause the sign of the resulting value to differ from the sign of the input value.

另见this thread.