Java位操作长-删除一些位

Question

I have a long number. Now, what I want to do is the following:

long l = "001000......10001100000" (bits representation)

I need to remove the 3 ^rd and 4 ^th bits of it ( ie 10) from the left and convert the remaining 62 bits to long

long newl = "0000......10001100000" (bits representation)

Can anybody help me do this efficiently in Java ?

Answer 1

You typically set bits by OR ing them with a mask; you clear bits by AND ing them with the complement of a mask:

long mask = 3L << 60; // 001100...
long newl = l & ~mask;     // Clear the bits in the mask.

If you're looking to remove the bits (and effectively shorten the value to 62 bits), then you can try this:

long topBits = ((3L << 62) & l) >> 2; // Top 2 bits, shifted right
long bottomBits = (~(15L << 60) & l); // Bottom 60 bits
long newl = topBits | bottomBits;

Answer 2

如果您将位列表作为字符串，则只需从该字符串中删除第3位和第4位，然后使用Long.parseLong(string, 2) 。