了解 Java 数据类型 [英] Understanding Java data types

问题描述

1) 为什么不允许以下赋值:

1) Why is the following assignment not allowed:

byte b = 0b11111111; // 8 bits or 1 byte

但允许此分配:

int i = 0b11111111111111111111111111111111; //32 bits or 4 bytes

两种类型都是有符号的，我希望 b 和 i 是 -1.

Both types are signed, and I would expect b and i were -1.

2) 为什么整数 MIN_VALUE 没有符号?

2) Why doesn't the Integer MIN_VALUE have a sign?

public static final int   MIN_VALUE = 0x80000000;

但是字节 MIN_VALUE 确实有符号?

but the Byte MIN_VALUE does have a sign?

public static final byte   MIN_VALUE = -128;

推荐答案

所有整数字面量都具有 int 类型(除非后缀为 L 或 l).因此，在第一种情况下，您将 int 存储到 byte 中.没有强制转换，这样的缩小转换是不允许的，除非右边是一个常量，如果值在范围内是允许的，即 -128 到 127.0b11111111 是 255，但不在范围内.

All integer literals have type int (unless suffixed by an L or l). Thus, in the first case, you're storing an int into a byte. A narrowing conversion like this is not allowed without a cast, except that if the right side is a constant, it's allowed if the value is in range, which is -128 to 127. 0b11111111 is 255, though, which is not in range.

至于为什么 int i = 0b1111111111111111111111111111111 是允许的:这几乎是因为 JLS 是这么说的".事实上，该特定示例出现在 JLS 3.10.1.有一个规则是 decimal int 类型的文字不能超过 214743647(特定情况除外 -2147483648)，但没有关于二进制文字的规则除了它们必须适合 32 位.

As for why int i = 0b11111111111111111111111111111111 is allowed: it's pretty much "because the JLS says so". In fact, that specific example appears in JLS 3.10.1. There's a rule that decimal literals of type int cannot exceed 214743647 (except in the specific case -2147483648), but there's no rule about binary literals except that they have to fit into 32 bits.

正如我在评论中提到的，第二个问题实际上是关于编写代码的程序员的风格偏好的问题，无法回答.

As I mentioned in a comment, the second question is really a question about the style preference of the programmers who wrote the code, and it's impossible to answer.

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