CODEWITHSUNDEEP
javabit-manipulationbit
Cratylus
Cratylus

Reputation: 54094

Why does it get the single bit like that?

I am looking some part of code that is supposed to get a single bit from an int.
It is as follows: 

private int getBit( int token, int pos){  
   return ( token & ( 1 << pos ) ) != 0 ? 1 : 0;   
}

My question is why doesn't it do it the following (simpler) way? 

return token & ( 1 << pos );

I expect that it will also return a 0 or 1.
Am I wrong on this? Is the second (mine) version wrong?

Upvotes: 0

Views: 84

Answers (4)

Hot Licks
Hot Licks

Reputation: 47739

Of course, you can use something on the order of

(token >> pos) & 1

Upvotes: 2

murgatroid99
murgatroid99

Reputation: 20297

Your version is wrong. When you execute

return token & ( 1 << pos );

if it is non-zero, then you get an int with every bit except the pos bit zeroed out, because that is the number on the right side of the & operator. This obviously would only be 1 if pos==0.

This happens because the & operator simply takes the bitwise and operation between corresponding bits in the two ints. Since 1 << pos has a 1 bit in a position besides the lowest and token can presumably be any int, the result can also have a 1 bit in a position other than the lowest, making it greater than 1.

Upvotes: 2

Jerome
Jerome

Reputation: 1519

Your version does not return the value of the bit at position pos. It returns the value 0 or 2^(pos-1).

Upvotes: 3

Mat
Mat

Reputation: 206841

Your version returns 0 or 1<<pos.

That won't matter if it is used in boolean context. But it might otherwise.

Upvotes: 2

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