bitwise operations

using bitwise operations how could i split a 16 digit binary number into two seperate 8 digit binary numbers?

like this:

Code:

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int number_16bit = 0xFFFF;
int number1_8bit = (number_16bit >> 8) & 0xFF;
int number2_8bit = number_16bit & 0xFF;

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number2 will contain the least significant bits (right bits) and number1 will contain the leftmost bits. Since number1/number2 are only 8 bits, you can also use a char type to store the content.

Quote:

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Originally Posted by KONI

like this:

Code:

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int number_16bit = 0xFFFF;
int number1_8bit = (number_16bit >> 8) & 0xFF;
int number2_8bit = number_16bit & 0xFF;

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number2 will contain the least significant bits (right bits) and number1 will contain the leftmost bits. Since number1/number2 are only 8 bits, you can also use a char type to store the content.

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in that case will number2 the one that contains the least significant bits always contain 1111 1111?
i obtained the left side of the 16 digit binary number by doing the following

Code:

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        int value2 = FileSize >> 8;

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I am then going to conver the integer values into char values.

Number2 will contain the least significant bits of your input since I masked it with an 0xFF mask. The bitwise AND together with the mask (0xFF) keeps the 8 least significant bits without changing their values and sets the rest to 0.

Code:

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int value2 = FileSize >> 8;

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The code above only works if FileSize is exactly 16 bits, it won't work if FileSize is 17 bits or above. By masking with an 8 bit mask, you're making sure that the rest of the bits is discarded/set to zero.

Quote:

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Originally Posted by KONI

Number2 will contain the least significant bits of your input since I masked it with an 0xFF mask. The bitwise AND together with the mask (0xFF) keeps the 8 least significant bits without changing their values and sets the rest to 0.

Code:

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int value2 = FileSize >> 8;

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The code above only works if FileSize is exactly 16 bits, it won't work if FileSize is 17 bits or above. By masking with an 8 bit mask, you're making sure that the rest of the bits is discarded/set to zero.

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I know it will always be exactly 16 bits as i put the data in as 16 bits, thanks a lot for your help, anding a binary number with that mask only checks the first 8 digits and then set the rest to 0?
so if i and the digits 1 0 1 1 1 1 1 1 0 1 0 1 1 1 1 0 with 0xFF the output should be 0 1 0 1 1 1 1 0.

exactly, this is because the mask 0xFF has every bit set to 0 if you bitwise AND it with a number larger than 8 bits. And because, if you AND whatever bit with 0, the result is 0, you eliminate all the bits apart from the 8 first ones.

But for joining them back together as one int would i have too loop around each individual bit, for both of the seperate numbers, comparing them against a 16 bit 0 number?

Or you could just shift them in together in two swppos. Kind of the opposite of what you did to extract them in the first place.

Quote:

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Originally Posted by black_watch

But for joining them back together as one int would i have too loop around each individual bit, for both of the seperate numbers, comparing them against a 16 bit 0 number?

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Like this:

Code:

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number_16bit = (number1_8bit << 8) | number2_8bit;

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C makes sure that, if you left shift a number, only 0's will be added.

That worked great, thanks a lot guys, i think im understanding this whole bitwise thing a little better now.