Java 文字值赋值行为

Question

在 Kathy Sierra 的 SCJP 指南一书中，在作业一章中，我们了解到我们可以声明这样的东西 byte b = 7;。幕后代码是 byte b = (byte) 7;。之所以如此，是因为在 Java 中，数字 7 被视为文字 int 值，因此必须转换为 int。

现在其他情况。 Double 可以包括 float 值中包含的每个字节，因为它是更大的数据类型。那么我们可以说 float f = 10.543; 因为 10.543 是一个很小的值，应该适合一个 float 。此类数字的字面值也被视为 Double，因此编译器应将其隐式转换为 float 。但事实并非如此，编译器阻止了我们。我们必须在该值之后附加 F 或 f。

为什么字面值赋值会出现这两种相互冲突的行为？简而言之，如果 byte b = 7 是可能的。为什么 float f = 10.543 不可能？

Answer 1

您可以阅读 JLS 5.2 Assignment Conversion

The compile-time narrowing of constants means that code such as:

 byte theAnswer = 42;

is allowed. Without the narrowing, the fact that the integer literal 42 has type int would mean that a cast to byte would be required:

byte theAnswer = (byte)42;  // cast is permitted but not required

If the type of the expression cannot be converted to the type of the variable by a conversion permitted in an assignment context, then a compile-time error occurs.

If the type of the variable is float or double, then value set conversion (§5.1.13) is applied to the value v

JLS #3.10.2.Floating-Point Literals

A floating-point literal is of type float if it is suffixed with an ASCII letter F or f; otherwise its type is double and it can optionally be suffixed with an ASCII letter D or d

5.1.2. Widening Primitive Conversion

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.

我希望以上能澄清你的疑问。