What does this bit wise operation doing?

Question

I read about bitwise operations a lot, but still, I couldn't give a meaning to this line.

((text.flags & ~Text.BOLD) & ~Text.ITALIC) | Text.BOLD | Text.ITALIC

It seems like the author is trying to be sure that this text doesn't have styles BOLD and ITALIC, and then he makes the text ITALIC and BOLD.

Am I right, or missing some detail?

score 2 · Answer 1 · answered Mar 13 '17 at 00:14

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No, you've got it; the & operations are erasing the BOLD and ITALIC bits, while the | operations set them.

answered Mar 13 '17 at 00:14

Scott Hunter

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OneCricketeer · Accepted Answer · 2017-03-13T17:37:11.793

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It seems to be turning off all flags not BOLD and not ITALIC (via & with the complement), and then ensuring that BOLD | ITALIC is set (via |).

The end result would be that for any input text regardless of style, the output is text

Could be re-written as

int bold_italic = Text.BOLD | Text.ITALIC;
text.flags = (text.flags & ~bold_italic) | bold_italic;

edited Mar 13 '17 at 17:37

answered Mar 13 '17 at 00:23

OneCricketeer

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So as a result, it creates a text that has only BOLD and ITALIC styles. – Mar 13 '17 at 17:12
It seems so, yes. Certainly, you could test that yourself? – OneCricketeer Mar 13 '17 at 17:35
Or you could just write `text.flags |= bold_italic` – harold Mar 13 '17 at 17:44
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@Paulpro the other flags are not turned off anyway. `& ~bold_italics` turns off *bold_italics*. It's the same flags that are turned off and then turned on again. If that's not convincing by itself [here](http://haroldbot.nl/?q=%28textflags+%26+~bold_italic%29+%7C+bold_italic) is a step-by-step proof – harold Mar 13 '17 at 18:48
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Also, if the other flags *were* turned off, the whole thing would simplify to `text.flags = bold_italic` – harold Mar 13 '17 at 18:52
@harold You're right, my bad. This answer is wrong, it doesn't turn off all other flags. – Paul Mar 13 '17 at 18:52
@harold your haroldbot is great. I completely understand it now. – Mar 13 '17 at 23:56

score 2 · Answer 3 · edited Mar 13 '17 at 17:48

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Lets start with 4 bits flag.

BOLD = 0001; ITALIC = 0010

flags & ~BOLD =   
flags & ~0001 = 
flags & 1110 = clear BOLD flag.

flags | ITALIC = 
flags | 0010 = 
flags | 0010 = set ITALIC flag

edited Mar 13 '17 at 17:48

OneCricketeer

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answered Mar 13 '17 at 00:39

Nguyễn Hoàng Đông Phương

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~ operator is actually complement not NOT operator. It's very hard to follow your answer. – Mar 13 '17 at 17:11
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@AlwaysNewbie that's just a different name for the same thing – harold Mar 13 '17 at 17:45

What does this bit wise operation doing?

3 Answers3