I'm trying to know the states of different bits over a char
Let's say I've:
char a = 11 //00001011 in binary
How can I retrieve bit number 5 which is currently 0 and cast it to a bool variable? And how can I set it?
-
In programming I would normally expect you to count bits starting with the rightmost bit being bit zero, so that the nth bit has value 2ⁿ, although you accepted an answer that used 1-based indexing...Neil– Neil2012-10-18 14:35:45 +00:00Commented Oct 18, 2012 at 14:35
-
@Neil - A good point to be sure. I'm used to talking with individuals who think of bit positions as 1-8 instead of the 0-7 as they're usually noted in reference manuals. I made that assumption here as well, so I did clarify that.Mike– Mike2012-10-18 14:51:39 +00:00Commented Oct 18, 2012 at 14:51
Add a comment
|
7 Answers
You can use some bitwise operators:
bool fifthBitSet = a & 0x10;
Or, more generally:
bool nThBitSet = a & ( 0x1 << n );
Missed the setting part:
a |= 0x1 << n; // set n'th bit
2 Comments
Neil
He said the fifth bit is zero, so it's definitely not
0x8 however you count it.
Luchian Grigore
@Neil my bad, I thought he meant from the left.
As you've noted:
char a = 11 // == 00001011 in binary
Since you said the bit number 5 is clear right now, let's *define the position of the bits as you're talking about them:
pos 8 7 6 5 4 3 2 1
value [0][0][0][0][1][0][1][1]
You want to test the 5th bit, that means you need a mask that accesses the fifth bit, here's some masks:
your number: 00001011
0x1 mask: 00000001
0x2 mask: 00000010
0x4 mask: 00000100
0x8 mask: 00001000
0x10 mask: 00010000 <-- So that's the one we want, the 5th bit
To set your bit you need a bit-wise OR (|)
a |= 0x10; // 00001011
// | 00010000 because 1 | anything = 1
// ----------
// 00011011 The value is set!
To test the bit we can use the bit-wise AND (&)
bool something = a & 0x10; // 00001011
// & 00010000 because only 1 & 1 = 1
// ----------
// 00000000 something will be 0 (false)
*it was pointed out that this is a 1-based indexing as opposed to the more typical 0-based indexing. Which is true. Based on the description I chose to read the question as "bit number 5" meaning typical (human) based counting of 1, 2, 3, 4, etc out of 8-bits.
If you'd rather use a 0-based indexing, this is no problem, just "shift" the same logic by one, the mask for the "5th" bit in 0-based indexing is 0x20
When talking about bits it's good to note which is the least significant bit, and 0 or 1 based to be totally clear.
Comments
Counting from 0 being the rightmost bit:
To test it:
int b = 5;
bool n = a & (1 << b);
To set (or clear) bit b:
if (n) { // set
a |= (1 << b); // RHS only has bit 'b' set
} else { // clear
a &= ~(1 << b); // RHS has every bit _except_ bit 'b' set
}
Comments
You mention bit number 5, I believe you're looking at the 5th least-significant bit.
You've also asked about how to set a bit at a specific position. For that, use bitwise OR - |:
a = a | 0x10;
As has already been answered, test the bit using bitwise AND - &:
bool isSet = a & 0x10;
Often you'll use a positional flag like 0x10, but the generalizations given using bit-shifts are very useful, especially if you wrap these into a function:
int pos = 5;
// set bit 5:
a = a | (0x1 << pos);
// test bit 5:
bool isSet = a & (0x1 << pos);
For more information, the wikipedia article on bitwise operation is pretty good.
Comments
You can test the bits with bitmasks:
a & 0b00010000
Comments
Well, you can use this set of functions to get and set the bit states, I suppose:
bool getBit(int data, int bitNumber) {
int flag = 1 << bitNumber - 1;
return (bool)(data & flag);
}
void setBit(int& data, int bitNumber) {
int flag = 1 << bitNumber - 1;
data |= flag;
}
...
int main() {
int a = 11;
cout << getBit(a, 5); // false
setBit(a, 5);
cout << getBit(a, 5); // true
}
Still I wonder why do you need to use this, and not bitsets. )
Comments
You can use bitwise AND to know the states.
unsigned int number = 11;
AND the number with the value in which the bit position you want to check alone is set and the remaining bits to 0.
Suppose I want to check bit 0, then
if(a&0x00000001)
puts(" 01h bit position is 1")
else
puts(" 01h bit position is 0")
