Question

I have a simple question – I need to write a function for my program to change the 3rd bit of a given byte.

I wrote those lines :

public byte turnOn(Byte value)
{
    int flag = 8;
    value = (byte) (value | flag);
    return value;
}

I’m not sure if it’s the right way to do that, because I saw also this way (with which I am unfamiliar)

value = (byte) (value | (1 << 2) );

which way is better, and what does 1 << 2 means (2 means the third bit, but what is the 1 )

Thanks!

Answer 1

1 << 2 means 1 shifted two bits to the left. Since shifting left by one bit is similar to multiplying by two, this gives 4. In binary, this is

00000100

i.e. the 3rd bit from the right is set.

The constant 1 is used since that number only has a single bit set – the rightmost bit. After shifting left, only the 3rd bit (from the right) is set:

00000001 original value
00000010 after shifting left once
00000100 after shifting left again

I prefer using 1 << 2 instead of a constant like 8, as it makes it clearer which bit is being set. It also prevents you inadvertently using a constant that has multiple bits set – unless you actually want that, of course. Even then, it’s clearer in my opinion to add together several bits, for clarity:

final int bitsToSet = (1 << 2) + (1 << 5);