by Oscar
When I strated out, I was really confused about bit shift andhexadecimal number and binary. This is a good example how confusing it could get sometimes.Occasionallyyou see this in the code, for exmaple in C++:
byte a = ( b >> 8) & 0xff; // b is an integer
Okay, I know what “>> 8” means, we are shifting variable “b” to the rightby 8 bits. But, what the heck is “& 0xff” doing there?
Well, 0xff is the hexadecimal number FF which has a integer value of 255. And the binary representation of FF is 00000000000000000000000011111111 (under the 32-bit integer).
The & operator performs a bitwise AND operation. a & b will give you an integer with a bit pattern that has a 0 in all positions where b has a 0, while in all positions where b has a 1, the corresponding bit value froma is used (this also goes the other way around). For example, the bitwise AND of 10110111 and00001101 is 00000101.
In a nutshell, “& 0xff”effectively masks the variable so it leaves only the value in the last 8 bits, and ignores all the rest of the bits.
It’s seen most in cases like when trying to transform color values from a special format to standard RGB values (which is 8 bits long).
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6 comments
Shayan 14th April 2018 - 11:51 pm
I was confused about 0xff. this article helped me alot. thanks for sharing.
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Hoswar Ruby 5th January 2017 - 5:01 am
i think , it’s really wierd …
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Muslim Aswaja 15th March 2016 - 3:58 pm
Thank you, Oscar! It’s really helpfull for me.
:-)
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Simon 18th November 2015 - 9:39 pm
Hi Oscar,
Thanks for your knowledge, it is really helpful to me!
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Abder-Rahman 23rd August 2015 - 5:59 pm
Thanks for the nice tutorial!
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bit_shift 2nd December 2014 - 9:47 pm
Forgive me, but isn’t it >> a right shift, because you know it point right and move the bits to the right, and << is the left shift?
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