if (index & 0x88) == 0, how this works?

Question

Can anyone show me how this operation works? Index is number and it can be any number from 0 to 128. I just don't understand how (index & 0x88) can be 0 or not.

Any help will be greatly appreciated!

0x88, when written in binary, is 0b10001000. Therefore, `index & 0x88` is zero precisely when both the 8th and 4th bits of index are 0. Usually this type of condition is used when you are testing for bit-based flags (in this case, the combination of flags indicated by the 8th and 4th flags being unset.) — dlev, Apr 24 '13 at 05:36
See http://stackoverflow.com/questions/1746613/bitwise-operation-and-usage/1746642#1746642 — paxdiablo, Apr 24 '13 at 05:39

score 3 · Answer 1 · answered Apr 24 '13 at 05:37

3

0x88 is equivalent to 10001000 in binary. Thus, it will be 0 iff the binary value of the index is 0xxx0xxx, where x is any binary digit.

answered Apr 24 '13 at 05:37

Yuushi

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score 2 · Accepted Answer · answered Apr 24 '13 at 05:39

The & operator is a bitwise AND, if the binary digits of 0x88 and corresponding spot in index are both 1, it will not == 0. In the opposite case, if none of the digits are both 1, then the outcome of the & will be 0

In this case, your hex number 88 is 10001000 in binary, so (index & 10001000) can equal to 0 as long as index has 0 in it's 4th and 8th positions (for example, 01110111)

score 0 · Answer 3 · answered Apr 24 '13 at 05:37

0

& is the bitwise and operator (when applied to numbers).

For example, 110 & 101 = 100

0x88 is 10001000 in binary.

answered Apr 24 '13 at 05:37

Ofiris

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if (index & 0x88) == 0, how this works?

3 Answers3