Java char 是有符号的还是无符号的？

Question

Java char是一个16位的数据类型，但是在对其进行运算时，它是有符号的还是无符号的？

你可以在算术中将它用作无符号的 16 位整数吗？

例如，以下是正确的吗？

char c1;
char c2;

int i = c1 << 16 | c2;

或者是否需要先去除 c2 的符号扩展位？

（我确信这个问题的答案在别处，但明显的搜索似乎没有找到）。

Answer 1

来自specs

For char, from '\u0000' to '\uffff' inclusive, that is, from 0 to 65535

因为它是 16 位，这意味着它们是无符号的。

Answer 2

char 是无符号的。来自 JLS§4.2.1:

For char, from '\u0000' to '\uffff' inclusive, that is, from 0 to 65535

...但请注意，当您对它们使用任何 various mathematic operations 时（包括按位运算和移位运算），它们会根据其他操作数的类型扩展为另一种类型，并且其他类型可能已签名：

Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:

If either operand is of type double, the other is converted to double.

Otherwise, if either operand is of type float, the other is converted to float.

Otherwise, if either operand is of type long, the other is converted to long.

Otherwise, both operands are converted to type int.

例如，char + char是int，所以：

public class Example {
    public static void main(String[] args) {
        char a = 1;
        char b = 2;

        char c = a + b;          // error: incompatible types: possible lossy conversion from int to char
        System.out.println(c);
    }
}

重新位扩展，如果我们按照the link above来加宽原语转换：

A widening conversion of a char to an integral type T zero-extends the representation of the char value to fill the wider format.

所以 char 0xFFFF 变成 int 0x0000FFFF，而不是 0xFFFFFFFF。

Java char 是有符号的还是无符号的？

Is Java char signed or unsigned for arithmetic?

java

char

unsigned