Let's take int
as an example:
int SetBitWithinRange(const unsigned from, const unsigned to)
{
//To be implemented
}
SetBitWithinRange
is supposed to return an int
in which all and only the bits starting at bit from
to bit to
are set, when from is smaller
than to
and both are in the range of 0
to 32
.
e.g.:
int i = SetBitWithinRange(2,4)
will result in i
having the value of 0b00...01100
Setting all bits can be done by using the | (OR) bit operator with 1s for each of the bits. This is because 1 OR with any number sets the number as 1.
Setting a bit Use the bitwise OR operator ( | ) to set a bit.
We can toggle all bits of a number using NOT (~) bitwise operation. Doing NOT (~) operation on a number toggles all of its bits.
Here are some ways. First, some variants of "set n
bits, then shift by from
". I'll answer in C# though, I'm more familiar with it than I am with C. Should be easy to convert.
uint nbits = 0xFFFFFFFFu >> -(to - from);
return nbits << from;
Downside: can't handle an empty range, ie the case where to <= from
.
uint nbits = ~(0xFFFFFFFFu << (to - from));
return nbits << from;
Upside: can handle the case where to = from
in which case it will set no bits.
Downside: can't handle the full range, ie setting all bits.
It should be obvious how these work.
Alternatively, you can use the "subtract two powers of two" trick,
(1u << to) - (1u << from)
Downside: to
can not be 32, so you can never set the top bit.
Works like this:
01000000
^^^^^^ "to" zeroes
100
^^ "from zeroes"
-------- -
00111100
To the right of the 1 in the "from" part, it's just zeroes being subtracted from zeroes. Then at the 1 in the "from" part, you will either subtract from a 1 (if to == from
) and get 0 as a result, or you'll subtract a 1 from a 0 and borrow all the way to the 1 in the to
part, which will be reset.
All true bitwise methods that have been proposed at the time of writing have one of those downsides, which raises the question: can it be done without downsides?
The answer is, unfortunately, disappointing. It can be done without downsides, but only by
To give an example of 1, you can just pick any of the previous methods and add a special case (with an if
or ternary operator) to work around their downside.
To give an example of 2: (not tested)
uint uppermask = (((uint)to >> 5) ^ 1) << to;
return uppermask - (1u << from);
The uppermask
either takes a 1 and shifts it left by to
(as usual), or it takes a 0 and shifts it left (by an amount that doesn't matter, since it's 0 that's being shifted), if to == 32
. But it's kind of weird and uses more operations.
To give an example of 3, shifts that give zero when you shift by the operand size or more would solve this very easily. Unfortunately, that kind of shift isn't too common.
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