Java's clever method of displacement operation

Left shift operation: If x is of type byte, short, or char, then x is promoted to int;

2) If x is of type byte, short, char, or int, then n is reassigned (the process is: take the lower 5 of n's complement bits are then converted into decimal int values, which is equivalent to taking n modulo 32: n=n%32);

If x is long type, n is reassigned (the process is: take the lower 6 bits of n's complement and then Converting to a decimal int value is equivalent to taking n modulo 64: n=n%64);

(Because the int type is 4 bytes, that is, 32 bits, moving 32 bits will make no sense. For long, it is modulo 64)

3) Shift x to the left by n digits, and the entire expression produces a new value (the value of x remains unchanged);

5cfffb9747eed5aa7c7d14042e38a51a> b", first perform b mod 32||64 operation, If a is of type int, take mod 32. If a is of type double, take mod 64, and then use the general shift operation rules mentioned above to shift.

At this point, you can understand why there is this statement

　// 符号位为1的右移   
// 最高4位为1000， 右移1位，变成1100也就是c,   
for (int i = 0;i < 8 ;i++)   
System.out.print( Integer.toHexString(0x80000000 >> i) + " ");

in the BitSet class, because Programmers who are familiar with jdk know that writing 1L << (bitIndex % 64) is redundant for jdk.

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