How are bitwise operators being used in this code?

Question

I was looking at PSPDFkit sample code and saw this:

NSDictionary *options = @{kPSPDFProcessorAnnotationTypes :
                               @(PSPDFAnnotationTypeNone & ~PSPDFAnnotationTypeLink)
                         };

The constants PSPDFAnnotationTypeNone and PSPDFAnnotationTypeLink are defined below:

// Available keys for options. kPSPDFProcessorAnnotationDict in
// form of pageIndex -> annotations.
// ..
extern NSString *const kPSPDFProcessorAnnotationTypes;

// Annotations defined after the PDF standard.
typedef NS_OPTIONS(NSUInteger, PSPDFAnnotationType) {
    PSPDFAnnotationTypeNone      = 0,
    PSPDFAnnotationTypeLink      = 1 << 1,  // Links and multimedia extensions
    PSPDFAnnotationTypeHighlight = 1 << 2,  // (Highlight, Underline, StrikeOut) - 
    PSPDFAnnotationTypeText      = 1 << 3,  // FreeText
    PSPDFAnnotationTypeInk       = 1 << 4,
    PSPDFAnnotationTypeShape     = 1 << 5,  // Square, Circle
    PSPDFAnnotationTypeLine      = 1 << 6,
    PSPDFAnnotationTypeNote      = 1 << 7,
    PSPDFAnnotationTypeStamp     = 1 << 8,
    PSPDFAnnotationTypeRichMedia = 1 << 10, // Embedded PDF videos
    PSPDFAnnotationTypeScreen    = 1 << 11, // Embedded PDF videos
    PSPDFAnnotationTypeUndefined = 1 << 31, // any annotation whose type not recognized
    PSPDFAnnotationTypeAll       = UINT_MAX
};

I understand that ~ is the bitwise not operator and & the bitwise and operator, but what is the purpose of their application in this code?

NSDictionary *options = @{kPSPDFProcessorAnnotationTypes :
                               @(PSPDFAnnotationTypeNone & ~PSPDFAnnotationTypeLink)
                         };

Based on comments below, the above could have been written simply as

NSDictionary *options = @{kPSPDFProcessorAnnotationTypes :@(PSPDFAnnotationTypeNone)};

Since it is the same as (0 & ~2) => 0. What's the point of adding the & ~PSPDFAnnotationTypeLink part?

In the context of this code it seems to be an obfuscated way of getting 0: `(0 & ~2)` => 0 — Carl Veazey, Apr 17 '13 at 06:38
so what value does writing the extra `& ~2` add? why not just put `0`? — abbood, Apr 17 '13 at 06:39
That I don't know... would vote to reopen were question to focus on why. — Carl Veazey, Apr 17 '13 at 06:47
@CarlVeazey cool.. updated the question and voted for it to re-open — abbood, Apr 17 '13 at 06:58
Why is the `enum` skipping 1<<0? Or 1<<9? And why is it using the wrong type for the `All` option? — jscs, Apr 17 '13 at 07:08
John: The 1<<0 is used in more recent versions, 1<<9 has legacy reasons, and what is wrong with UINT_MAX? (Since the enum type is NSUInteger?) — steipete, Apr 17 '13 at 21:42
@steipete: `NSUInteger`'s underlying type is `unsigned long`, and the maximum is `NSUIntegerMax`. An `NSUInteger` will be longer than an `unsigned int` on the desktop or if iOS ever gets 64-bit processors. — jscs, Apr 18 '13 at 18:46
@steipete: Oh! I didn't realize this was your code; I thought it was the question poster's. Well, glad I could help! — jscs, Apr 19 '13 at 20:52

yinkou · Accepted Answer · 2013-04-17T06:45:35.707

2

"~" is the bitwise not-operator.

As "&" the bitwise and.

These are usually used for bitmask (like in your example) or other binary operations (as the name lets suggest). More info on wiki - Operators in C and C++.

They are in no relationship to literals.

edited Apr 17 '13 at 06:45

answered Apr 17 '13 at 06:30

yinkou

5,756
2
24
40

2

Replace "binary" with "bitwise". – rmaddy Apr 17 '13 at 06:33

score 0 · Answer 2 · answered Apr 17 '13 at 06:32

0

First of all, I don't know obj-c, only C, but I guess the '&' is 'bitwise AND' and the '~' is bitwise NOT.

answered Apr 17 '13 at 06:32

Péter Trombitás

39
2

score 0 · Answer 3 · edited May 23 '17 at 12:18

0

It's the bitwise NOT operator (same as many C-based languages), which inverts all bits in the underlying value.

So, for example, the eight bit value 0x57 (binary 0101 0111) becomes 1010 1000 or 0xa8.

See here for a more complete description of the various bitwise operators.

edited May 23 '17 at 12:18

Community

1
1

answered Apr 17 '13 at 06:32

paxdiablo

854,327
234
1,573
1,953

How are bitwise operators being used in this code?

3 Answers3