为什么显式类型转换需要从double到float而不是从int到byte？ [英] why explicit type casting required from double to float but not from int to byte?

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 of type 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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