If I have a byte, how would the method look to retrieve a bit at a certain position?
Here is what I have know, and I don't think it works.
public byte getBit(int position) {
return (byte) (ID >> (position - 1));
}
where ID
is the name of the byte I am retrieving information from.
public byte getBit(int position)
{
return (ID >> position) & 1;
}
Right shifting ID by position will make bit #position be in the furthest spot to the right in the number. Combining that with the bitwise AND &
with 1 will tell you if the bit is set.
position = 2
ID = 5 = 0000 0101 (in binary)
ID >> position = 0000 0001
0000 0001 & 0000 0001( 1 in binary ) = 1, because the furthest right bit is set.
You want to make a bit mask and do bitwise and. That will end up looking very close to what you have -- use shift to set the appropriate bit, use &
to do a bitwise op.
So
return ((byte)ID) & (0x01 << pos) ;
where pos
has to range between 0 and 7. If you have the least significant bit as "bit 1" then you need your -1
but I'd recommend against it -- that kind of change of position is always a source of errors for me.
to get the nth bit in integer
return ((num >> (n-1)) & 1);
In Java the following works fine:
if (value << ~x < 0) {
// xth bit set
} else {
// xth bit not set
}
value
and x
can be int
or long
(and don't need to be the same).
Word of caution for non-Java programmers: the preceding expression works in Java because in that language the bit shift operators apply only to the 5 (or 6, in case of long
) lowest bits of the right hand side operand. This implicitly translates the expression to value << (~x & 31)
(or value << (~x & 63)
if value
is long
).
Javascript: it also works in javascript (like java, only the lowest 5 bits of shift count are applied). In javascript any number
is 32-bit.
Particularly in C, negative shift count invokes undefined behavior, so this test won't necessarily work (though it may, depending on your particular combination of compiler/processor).