问题描述
考虑以下声明：
byte by = 5; //工作正常
字面数'5'的类型为int，小到足以容纳byte类型的变量。编译器在这里执行隐式类型转换（从int到byte）。
literal '5' is of type int and small enough to fit into a variable of type byte. Compiler does the implicit type casting here (from int to byte).
现在考虑以下场景：
float fl = 5.5; //compilation error
文字'5.5'的类型为double，也足够小以适应变量
类型的浮点数。为什么我们需要像这样显式地输入：
literal '5.5' is of type double, also small enough to fit into a variable oftype float. Why do we need to explicitly type cast like this:
float fl = (float) 5.5; //works fine
为什么编译器在浮点数的情况下不为我们进行转换？
Why compiler is not doing the casting for us in case of floating points?
推荐答案
在整数版本中，编译器知道 all 数字 5 可以存储在字节中。没有信息丢失。浮点值并不总是如此。例如， 0.1f 不等于 0.1d 。
In the integer version, the compiler knows that all the data in the number 5 can be stored in a byte. No information is lost. That's not always true for floating point values. For example, 0.1f isn't equal to 0.1d.
现在举例说明，你已经给出了小数值5.5 在 float 和 double ，所以你可以争辩说，在这种情况下，没有信息丢失 - 但语言规范必须使其有效是非常奇怪的：
Now for the example, you've given, the decimal value 5.5 is exactly represented in both float and double, so you could argue that in that case, no information is lost - but it would be pretty odd for the language specification to have to make this valid:
float f = 5.5;
但是这个无效：
float f = 5.6;
语言规范很乐意谈论一个数字是否符合 float / double （尽管 并不像你想象的那么简单）但是当它出现时是否可以准确表示文字，我认为永远不会详细说明。
The language specification is happy to talk about whether a number fits within the range of float/double (although even that isn't as simple as you might expect) but when it comes to whether a literal can be exactly represented, I don't think it ever goes into detail.
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