java有人能解释一下这种移位/长时间修补输出吗？

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

该值被视为int，可以这样模拟：

运行3：

public static void main(String[] args) {
    System.out.println("shiftwise Example A = " + (0x47494638 << 32));

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

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

输出：

shiftwise Example A = 1195984440
shiftwise Example B = 1195984440

Answer 2

如果未指定L后缀，则常量文字值将被视为int，并且§15.19 of the Java Language Specification中的此文本适用于：

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，您的左移位32位被转换为零位的移位，即没有变化