Java隐式转换

Question

使用以下代码:

Float a = 1.2;

有一个错误，因为它将小数作为 double 值，而 double 是比 float 更大的数据类型。

现在，它采用整数作为默认的 int 类型。那么，为什么下面的代码没有给出任何错误？

Byte b = 20;

Answer 1

编译器足够聪明，可以计算出 20 的位表示(int 值)可以装入 byte 而不会丢失数据。来自Java Language Specification §5.1.3 :

A narrowing primitive conversion from double to float is governed by the IEEE 754 rounding rules (§4.2.4). 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 .