7

I do not not know how to implement the following algorithm.

For example I have int=26, this is "11010" in binary. Now I need to implement one operation for 1, another for 0, from left to right, till the end of byte. But I really have no idea how to implement this. Maybe I can convert binary to char array, but I do not know how. btw, int equals 26 only in the example, in the application it will be random.

Brooks Moses
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revolt
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7 Answers7

9

Since you want to move from 'left to right':

unsigned char val = 26;  // or whatever

unsigned int mask;

for (mask = 0x80; mask != 0; mask >>= 1) {

    if (val & mask) {
        // bit is 1
    }
    else {
        // bit is 0
    }
}

The for loop just walks thorough each bit in a byte, from the most significant bit to the least.

Michael Burr
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  • I can't understand why val is unsigned char? – revolt Dec 16 '10 at 23:59
  • And mask, I dont know, how long will be this value. I generate it random. – revolt Dec 17 '10 at 00:02
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    I used `unsigned char` because you indicated you were interested in performing an operation on each bit "till the end of byte". If you're interested in an object that's larger than a byte then use the appropriate type, but you'll need to change the starting bit in the `mask`. Note: `mask` is *not* random - it tracks the bit being examined at each step. If you want the value you're checking the bits of to be random, that's fine. Set it to whatever, however, you want. I used `26` in the example because that's what you used in your example. Set to whatever data you need for your application. – Michael Burr Dec 17 '10 at 00:22
  • You said you wanted to go through each bit of a byte; a char is one byte. If you want to go through each bit of an int, you could do that instead (but using a different start mask, as prev. comment says). – comingstorm Dec 17 '10 at 01:02
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    It might be a bit more self-documenting to initialize `mask=1<<7` instead of `mask=0x80`. Comparably, for an int, you would start out with `mask=1<<31`, to begin at the most significant bit, or `mask=1<<7` if you want to begin at the most significant bit of the least significant byte. – comingstorm Dec 17 '10 at 01:05
5

I use this option:

isBitSet = ((bits & 1) == 1);
bits = bits >> 1

I find the answer also in stackoverflow:

How do I properly loop through and print bits of an Int, Long, Float, or BigInteger?

Community
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Barak Rosenfeld
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4

You can use modulo arithmetic or bitmasking to get what you need.

Modulo arithmetic:

int x = 0b100101;
// First bit
(x >> 0) % 2; // 1
// Second bit
(x >> 1) % 2; // 0
// Third bit
(x >> 2) % 2; // 1
...
etc.

Bitmasking

int x = 0b100101;
int mask = 0x01;
// First bit 
((mask << 0) & x) ? 1 : 0
// Second bit
((mask << 1) & x) ? 1 : 0
...
etc.
Swiss
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3

In C, C++, and similarly-syntaxed languages, you can determine if the right-most bit in an integer i is 1 or 0 by examining whether i & 1 is nonzero or zero. (Note that that's a single & signifying a bitwise AND operation, not a && signifying logical AND.) For the second-to-the-right bit, you check i & 2; for the third you check i & 4, and so on by powers of two.

More generally, to determine if the bit that's jth from the right is zero, you can check whether i & (1 << (j-1)) != 0. The << indicates a left-shift; 1 << (j-1) is essentially equivalent to 2j-1.

Thus, for a 32-bit integer, your loop would look something like this:

unsigned int i = 26;  /* Replace this with however it's actually defined.  */

int j;
for (j = 31; j >= 0; j--)
{
  if ((i & (1 << (j-1))) != 0)
    /* do something for jth bit is 1 */
  else
    /* do something for jth bit is 0 */
}

Hopefully, that's enough to get you started.

Brooks Moses
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  • It's not the constant 26, it was only for example. – revolt Dec 16 '10 at 21:34
  • I should have been clearer -- I was assigning "i = 26" in my code as an example following your example; I would assume you would define it some other way. I'll clarify. – Brooks Moses Dec 16 '10 at 21:37
3

Came across a similar problem so thought I'd share my solution. This is assuming your value is always one byte (8 bits)

Iterate over all 8 bits within the byte and check if that bit is set (you can do this by shifting the bit we are checking to the LSB position and masking it with 0x01)

int value = 26;
for (int i = 0; i < 8; i++) {
    if ((value >> i) & 0x01) {
        // Bit i is 1
        printf("%d is set\n", i);
    }
    else {
        // Bit i is 0
        printf("%d is cleared\n", i);
    }
}
xulo
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0

I'm not exactly sure what you say you want to do. You could probably use bitmasks to do any bit-manipulation in your byte, if that helps.

tobier
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-1

Hi Look up bit shifting and bitwise and.

Mikkel Lund
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