I would like to set the i-th bit to zero no matter what the i-th bit is.
unsigned char pt = 0b01100001;
pt[0] = 0; // its not how we do this...
Setting it to one, we can use a mask pt | (1 << i)
but i'm not sure how to create a mask for setting 0, if thats possible.
You just have to replace the logical OR with a logical AND operation. You would use the & operator for that: pt = pt & ~(1 << i); You have to invert your mask because logical AND ing with a 1 will maintain the bit while 0 will clear it... so you'd need to specify a 0 in the location that you want to clear.
Clearing a bit Use the bitwise AND operator ( & ) to clear a bit. number &= ~(1 << x); That will clear bit x . You must invert the bit string with the bitwise NOT operator ( ~ ), then AND it.
3. Which of the following bitwise operations will you use to set a particular bit to 0? Explanation: 1 AND 0 = 0, 0 AND 0 = 0, any bit AND with 0 gives 0.
bitset::flip() is a built-in STL in C++ which flips the bits. If no parameter is passed in the function, then it flips all the bit values converting zeros to ones and ones to zeros. If a parameter position is passed, it flips the bit at the position only.
You just have to replace the logical OR
with a logical AND
operation. You would use the &
operator for that:
pt = pt & ~(1 << i);
You have to invert your mask because logical AND
ing with a 1
will maintain the bit while 0
will clear it... so you'd need to specify a 0
in the location that you want to clear. Specifically, doing 1 << i
will give you a mask that is 000...010..000
where the 1
is in the bit position that you want, and inverting this will give 111...101...111
. Logical AND
ing with this will clear the bit that you want.
You could stick with this:
// Set bit at position `bitpos` in `pt` to `bitval`
unsigned char bitpos = 1;
unsigned char pt = 0b01100001;
bool bitval = 1;
// Clear the bit
pt &= ~(1u << bitpos);
// Set the bit
pt |= (bitval << bitpos);
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