How to set/unset a bit at specific position of a long?

Question

How to set/unset a bit at specific position of a long in Java ?

For example,

long l = 0b001100L ; // bit representation 

I want to set bit at position 2 and unset bit at position 3 thus corresponding long will be,

long l = 0b001010L ; // bit representation 

Can anybody help me how to do that ?

Answer 1

To set a bit, use:

x |= 0b1; // set LSB bit x |= 0b10; // set 2nd bit from LSB 

to erase a bit use:

x &= ~0b1; // unset LSB bit (if set) x &= ~0b10; // unset 2nd bit from LSB 

to toggle a bit use:

x ^= 0b1; 

Notice I use 0b?. You can also use any integer, eg:

x |= 4; // sets 3rd bit x |= 0x4; // sets 3rd bit x |= 0x10; // sets 9th bit 

However, it makes it harder to know which bit is being changed.

Using binary allows you to see which exact bits will be set/erased/toggled.

To dynamically set at bit, use:

x |= (1 << y); // set the yth bit from the LSB 

(1 << y) shifts the ...001 y places left, so you can move the set bit y places.

You can also set multiple bits at once:

x |= (1 << y) | (1 << z); // set the yth and zth bit from the LSB 

Or to unset:

x &= ~((1 << y) | (1 << z)); // unset yth and zth bit 

Or to toggle:

x ^= (1 << y) | (1 << z); // toggle yth and zth bit 

Answer 2

The least significant bit (lsb) is usually referred to as bit 0, so your 'position 2' is really 'bit 1'.

long x = 0b001100;  // x now = 0b001100 x |= (1<<1);        // x now = 0b001110 (bit 1 set) x &= ~(1<<2);       // x now = 0b001010 (bit 2 cleared)