I have an enum declaration using bit flags and I cant exactly figure out on how to use this.
enum
{
kWhite = 0,
kBlue = 1 << 0,
kRed = 1 << 1,
kYellow = 1 << 2,
kBrown = 1 << 3,
};
typedef char ColorType;
I suppose to store multiple colors in one colorType I should OR the bits together?
ColorType pinkColor = kWhite | kRed;
But suppose I would want to check if pinkColor contains kRed, how would I do this?
Anyone care to give me an example using the provided ColorType example ?
Yes, use bitwise OR (|) to set multiple flags:
ColorType pinkColor = kWhite | kRed;
Then use bitwise AND (&) to test if a flag is set:
if ( pinkColor & kRed )
{
// do something
}
The result of & has any bit set only if the same bit is set in both operands. Since the only bit in kRed is bit 1, the result will be 0 if the other operand doesn't have this bit set too.
If you need to get whether a particular flag is set as a BOOL rather than just testing it in an if condition immediately, compare the result of the bitwise AND to the tested bit:
BOOL hasRed = ((pinkColor & kRed) == kRed);
Related
I am facing a simple problem by i have found a solution to it :).
if we have for example this :
typedef NS_OPTIONS(NSUInteger, UIViewAutoresizing) {
UIViewAutoresizingNone = 0,
UIViewAutoresizingFlexibleLeftMargin = 1 << 0,
UIViewAutoresizingFlexibleWidth = 1 << 1,
UIViewAutoresizingFlexibleRightMargin = 1 << 2,
UIViewAutoresizingFlexibleTopMargin = 1 << 3,
UIViewAutoresizingFlexibleHeight = 1 << 4,
UIViewAutoresizingFlexibleBottomMargin = 1 << 5
};
and a property like this :
#property (nonatomic, assign) UIViewAutoresizing autoresizingMask;
and this :
self.autoresizingMask = UIViewAutoresizingFlexibleRightMargin | UIViewAutoresizingFlexibleWidth ;
The questions is : How to know the number of items (UIViewAutoresizing values) in the autoresizingMask property ? ( in my example i have 2)
There is the __builtin_popcount function, which usually translates to a single instruction on most modern hardware. It basically gives you the number of set bits in an integer.
It's not very common to count the number of options, usually you just ask for every option you are interested in separately.
In this case however you can just traverse all 32 (64) bits and check how many bits are set to 1 (naive solution).
For more advanced solutions see: How to count the number of set bits in a 32-bit integer?
Note this works only if every option is specified by exactly 1 bit. I still advice to check for every option separately.
if ((self.autoresizingMask & UIViewAutoresizingFlexibleRightMargin) != 0) {
//do something
}
There are two ways to do this.
You either check inclusion one by one, and then either keep track of which ones or how many.
Or, you create a set of masks that cover all combinations and test against those for an exact match.
(More boilerplate code, increasing with number of enum options)
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?
"~" 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.
First of all, I don't know obj-c, only C, but I guess the '&' is 'bitwise AND' and the '~' is bitwise NOT.
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.
I am trying to implement the following typedef
typedef NS_OPTIONS (NSInteger, MyCellCorners) {
MyCellCornerTopLeft,
MyCellCornerTopRight,
MyCellCornerBottomLeft,
MyCellCornerBottomRight,
};
and correctly assign a value with
MyCellCorners cellCorners = (MyCellCornerTopLeft | MyCellCornerTopRight);
when drawing my cell, how can I check which of the options match so I can correctly draw it.
Use bit masking:
typedef NS_OPTIONS (NSInteger, MyCellCorners) {
MyCellCornerTopLeft = 1 << 0,
MyCellCornerTopRight = 1 << 1,
MyCellCornerBottomLeft = 1 << 2,
MyCellCornerBottomRight = 1 << 3,
};
MyCellCorners cellCorners = MyCellCornerTopLeft | MyCellCornerTopRight;
if (cellCorners & MyCellCornerTopLeft) {
// top left corner set
}
if (etc...) {
}
The correct way to check for this value is to first bitwise AND the values and then check for equality to the required value.
MyCellCorners cellCorners = MyCellCornerTopLeft | MyCellCornerTopRight;
if ((cellCorners & MyCellCornerTopLeft) == MyCellCornerTopLeft) {
// top left corner set
}
The following reference explains why this is correct and provides other insights into enumerated types.
Reference: checking-for-a-value-in-a-bit-mask
I agree with NSWill. I recently had a similar issue with wrong comparison.
The right if statement should be:
if ((cellCorners & MyCellCornerTopLeft) == MyCellCornerTopLeft){
I am using bitflag to enable/disable some functionality in my code. I have following enum.
typedef enum function {
function1 = 1 << 0,
function2 = 1 << 1,
function3 = 1 << 2,
function4 = 1 << 3
};
I know that I need to use following code to check which flag has been enabled.
if((flags & function1) == function1)
{
// do some action
}
In my enum, number of flags is large and to check every flag, I need to have that many "if" condition checks. Is there any way which can minimize the number of if statements required? I am new to objective C and looking to implement it in objective C. Thanks in advance.
Yes, combine the flags via a bitwise or:
if (flags & (function1 | function2 | function3 | function4))
{
// any of the flags has been set
}
Moreover, to check for a particular flag you don't need the part == function1, flags & function1 is sufficient as it will evaluate either to zero or function1.
I'm working on a Objective-C program where I'm getting bitfields over the network, and need to set boolean variables based on those bits.
Currently I'm representing the bitfields as int's, and then using bit shifting, similar to this (all the self properties are BOOL):
typedef enum {
deleteFlagIndex = 0,
uploadFlagIndex = 1,
moveFlagIndex = 2,
renameFlagIndex = 3
} PrivilegeFlagIndex;
int userFlag = 5; //for example
// this would be YES
self.delete = ((userFlag & (1 << deleteFlagIndex)) == (1 << deleteFlagIndex));
// this would be NO
self.upload = ((userFlag & (1 << uploadFlagIndex)) == (1 << uploadFlagIndex));
And this works (to the best of my knowledge) but I'm wondering - is there a more efficient concise way to code all the bit twiddling using a fancy trick/hack? I ask because I'll be doing this for a lot of flags (more than 30).
I did realize I could use this method this as well:
self.move = ((userFlag & (1 << moveFlagIndex)) > 0)
...which does reduce the amount of typing, but I don't know if there's a good reason to not do it that way.
Edit: Revised to say concise rather than efficient - I wasn't worried about execution performance, but rather tips and best practices for doing this in a smart way.
Try:
typedef enum {
deleteFlag = 1 << 0,
uploadFlag = 1 << 1,
moveFlag = 1 << 2,
renameFlag = 1 << 3
} PrivilegeFlags;
Now you can combine them using | directly.
Usually, it suffices to check against 0:
if (userFlag & deleteFlag) {
// delete...
}
You may want to try to use bitfields and let the compiler do the optimization itself.