java - 有人可以解释这种 shiftwise/Long 修补输出吗？

Question

运行 1:

public static void main(String[] args) {

        System.out.println("shiftwise Example A = " + (0x47494638 << 32));

        long someNumber = 0x47494638;
        long otherNumber = someNumber << 32;

        System.out.println("shiftwise Example B = " + otherNumber);



    }

输出:
shiftwise 示例 A = 1195984440
shiftwise 示例 B = 5136714056324874240

运行 2:(我刚刚在示例 A 中指定了“L”):

public static void main(String[] args) {

        System.out.println("shiftwise Example A = " + (0x47494638L << 32));

        long someNumber = 0x47494638;
        long otherNumber = someNumber << 32;

        System.out.println("shiftwise Example B = " + otherNumber);



    }

输出:
shiftwise 示例 A = 5136714056324874240
shiftwise 示例 B = 5136714056324874240

Answer 1

如果你没有指定L后缀，那么常量字面量值将被认为是一个int，并且这个文本来自§15.19 of the Java Language Specification适用:

If the promoted type of the left-hand operand is int, then only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x1f (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive.

因此，您的左移 32 位将转换为零 位的移位，即没有变化。