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Bit masking in C - How to get the first bit of a byte?

I have:

int8_t byteFlag;

and I want to get its first bit? I think I probably need to use &

and >>

, but I'm not exactly sure how. Any help?

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5 answers 


int func(int8_t byteFlag, int whichBit)
{
    if (whichBit > 0 && whichBit <= 8)
        return (byteFlag & (1<<(whichBit-1)));
    else
        return 0;
}

func(byteFlag, 1)

Will now return 1 bit from LSB. You can pass 8

like whichBit

to get the 8th bit (MSB).

<<

- left shift operand. It will move the value 1

to the appropriate location, and then we have to perform an operation &

to get the value of that private bit in byteFlag

.



for func(75, 4) 

75         -> 0100 1011
1          -> 0000 0001
1 << (4-1) -> 0000 1000 //means 3 times shifting left

75 & (1 << (4 - 1))

will give us 1

.

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Just disguise the high beat

int8_t high_bit = byteFlag & (1 << 7); //either 1 or 0

Another trick as it is a signed int



if (byteFlag < 0) firstBitSet = true;

The latter works because of the two's complement representation of numbers. The high bit is set if the number is negative.

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You must use the and operator.

If by "first bit" you mean LSB:

int firstBit = byteFlag & 1;

If by "first bit" you mean MSB:

int firstBit = byteFlag >> (sizeof(byteFlag) * 8 - 1);
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int8_t bit_value = (byteFlag & (1U << bitPosition)) ? 1 : 0 ;
/* now it up to you to decide which bit is the "first".
   bitPosition = 0 is the minor bit. */
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The solution is given below. To get the first bit of a number, set bit = 1;

int bitvalue(int8_t num, int bit)
{
    if (bit > 0 && bit <= 8)
        return ( (num >> (bit-1)) & 1 );
    else
        return 0;
}
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