Given an index for current character, how can I determine the number of line that the selected character is at?
Given a CTLine how can I determine the number of characters in it?
For the first one:
int currentCharacterIndex = 12; // You define this.
CFArrayRef lines = CTFrameGetLines(frame);
int currentLine = 0;
for (CTLineRef line in lines) {
currentLine++;
CFRange range = CTLineGetStringRange(line);
if (currentCharacterIndex > range.location)
break;
}
// Current line is now the line that the currentCharacterIndex resides at
For the second one:
CFRange range = CTLineGetStringRange(line);
CFIndex length = range.length; // Number of characters
Can't be sure these work as I haven't tested them but it's worth a go.
Related
How can I add a trail of six 0 of decimal places to a number in string type?
I tried PadRight but didn't get the result that I wanted. For example:
125 ===to===> 125,000000
14,5 ===to===> 14,500000
so on..
Is there any function in VB.NET can do that or do I need to split the string then combine them?
I'm not completely sure what you're asking, but if you want to add six 0's to a string then I would make a function, something like this:
private static string AddZeroes(string myString)
{
//Makes an array that is 6 elements higher than the original string
string[] myStringArray = new string[myString.Length + 6];
//Sets the array to have all elements of myString inside of it
for (int i = 0; i < myString.Length; i++)
{
myStringArray[i] = Convert.ToString(myString[i]);
}
//Adds the six 0's to the array
for (int i = myString.Length; i < myString.Length + 6; i++)
{
myStringArray[i] = "0";
}
//Sets myString to an empty string, then adds all the contents of the myStringArray to it
myString = "";
for (int i = 0; i < myStringArray.Length; i++)
{
myString += myStringArray[i];
}
//Returns the string that should now have six 0's added to the end
return myString;
}
If you wanted a comma or decimal, it shouldn't be too hard to just add one right before the for loop that adds the 0's, and also make sure that the array is the length of the string + 7 instead, to fit that extra character.
Writing a basic program to count the number of words in a string. I've changed my original code to account for multiple spaces between words. By setting one variable to the current index and one variable to the previous index and comparing them, I can say "if this current index is a space, but the previous index contains something other than a space (basically saying a character), then increase the word count".
int main(int argc, const char * argv[]) {
#autoreleasepool {
//establishing the string that we'll be parsing through.
NSString * paragraph = #"This is a test paragraph and we will be testing out a string counter.";
//we're setting our counter that tracks the # of words to 0
int wordCount = 0;
/*by setting current to a blank space ABOVE the for loop, when the if statement first runs, it's comparing [paragraph characterAtIndex:i to a blank space. Once the loop runs through for the first time, the next value that current will have is characterAtIndex:0, while the if statement in the FOR loop will hold a value of characterAtIndex:1*/
char current = ' ';
for (int i=0; i< paragraph.length; i++) {
if ([paragraph characterAtIndex:i] == ' ' && (current != ' ')) {
wordCount++;
}
current = [paragraph characterAtIndex:i];
//after one iteration, current will be T and it will be comparing it to paragraph[1] which is h.
}
wordCount ++;
NSLog(#"%i", wordCount);
}
return 0;
}
I tried adding "or" statements to account for delimiters such as ";" "," and "." instead of just looking at a space. It didn't work...any idea what I can do, logically speaking, to account for anything that isn't a letter (but preferably just limiting it to these four delimiters - . , ; and space.
A standard way to solve these types of problems is to build a finite state machine, your code isn't quite one but its close.
Instead of thinking about comparing the previous and current characters think in terms of states - you can start with just two, in a word and not in a word.
Now for each state you consider what the current character implies in terms of actions and changes to the state. For example, if the state is not in a word and the current character is a letter then the action is increment word count and the next state is in a word.
In (Objective-)C you can build a simple finite state machine using an enum to give the states names and a case statement inside a loop. In pseudo-code this is something like:
typedef enum { NotInWord, InWord } State;
State currentState = NotInWord;
NSUInteger wordCount = 0;
for currentChar in sourceString
case currentState of
NotInWord:
if currentChar is word start character -- e.g. a letter
then
increment wordCount;
currentState = InWord;
InWord:
if currentChar is not a word character -- e.g. a letter
then
currentState = NotInWord;
end case
end for
The above is just a step from your original algorithm - recasting it in terms of states rather than the previous character.
Now if you want to get smarter you can add more states. For example how many words are there in "Karan's question"? Two. So you might want to allow a single apostrophe in a word. To handle that you can add a state AfterApostrophe whose logic is the same as the current InWord; and modify InWord logic to include if the current character is an apostrophe the next state is AfterApostrophe - that would allow one apostrophe in a word (or its end, which is also valid). Next you might want to consider hyphenated words, etc...
To test if a character is a particular type you have two easy choices:
If this is just an exercise and you are happy to stick with the ASCII range of characters there are functions such as isdigit(), isletter() etc.
If you want to handle full Unicode you can use the NSCharacterSet type with its pre-defined sets for letters, digits, etc.
See the documentation for both of the above choices.
HTH
I don't understand, You should be able to add or statements....
int main(void) {
char paragraph[] = "This is a test paragraph,EXTRAWORDHERE and we will be testing out a string.";
char current = ' ';
int i;
int wordCount = 0;
for (i = 0; i < sizeof(paragraph); i++){
if ((paragraph[i] == 32 || paragraph[i] == 44) && !(current == 32 || current == 44)){ //32 = ascii for space, 44 for comma
wordCount++;
}
current = paragraph[i];
}
wordCount++;
printf("%d\n",wordCount);
return 0;
}
I suppose it would be better to change the comparison of current from a not equal to into an equal to. Hopefully that helps.
Right now, I am able to get the text ranges of each line in a UITextView containing n lines by using cycling through the tokenizer with paragraph granularity. Unfortunately, that means my search algorithm for the m-th line in the text is of order n. Is there any easier way for me to find the range other than making my algorithm log n? The following is how I find my text range for now:
- (UITextRange *)textRangeOfLineAtIndex:(NSUInteger)index {
UITextPosition *position = self.beginningOfDocument;
NSUInteger lineCount = 0;
while([self comparePosition:self.endOfDocument toPosition:position] == NSOrderedDescending && lineCount < index) {
position = [self.tokenizer positionFromPosition:position toBoundary:UITextGranularityParagraph inDirection:UITextStorageDirectionForward];
++lineCount;
}
return [self rangeEnclosingPosition:position withGranularity:UITextGranularityParagraph inDirection:UITextStorageDirectionForward];
}
I'm using a for-loop to determine whether the long double is an int. I have it set up that the for loop loops another long double that is between 2 and final^1/2. Final is a loop I have set up that is basically 2 to the power of 2-10 minus 1. I am then checking if final is an integer. My question is how can I get only the final values that are integers?
My explanation may have been a bit confusing so here is my entire loop code. BTW I am using long doubles because I plan on increasing these numbers very largely.
for (long double ld = 1; ld<10; ld++) {
long double final = powl(2, ld) - 1;
//Would return e.g. 1, 3, 7, 15, 31, 63...etc.
for (long double pD = 2; pD <= powl(final, 0.5); pD++) {
//Create new long double
long double newFinal = final / pD;
//Check if new long double is int
long int intPart = (long int)newFinal;
long double newLong = newFinal - intPart;
if (newLong == 0) {
NSLog(#"Integer");
//Return only the final ints?
}
}
}
Just cast it to an int and subtract it from itself?
long double d;
//assign a value to d
int i = (int)d;
if((double)(d - i) == 0) {
//d has no fractional part
}
As a note... because of the way floating point math works in programming, this == check isn't necessarily the best thing to do. Better would be to decide on a certain level of tolerance, and check whether d was within that tolerance.
For example:
if(fabs((double)(d - i)) < 0.000001) {
//d's fractional part is close enough to 0 for your purposes
}
You can also use long long int and long double to accomplish the same thing. Just be sure you're using the right absolute value function for whatever type you're using:
fabsf(float)
fabs(double)
fabsl(long double)
EDIT... Based on clarification of the actual problem... it seems you're just trying to figure out how to return a collection from a method.
-(NSMutableArray*)yourMethodName {
NSMutableArray *retnArr = [NSMutableArray array];
for(/*some loop logic*/) {
// logic to determine if the number is an int
if(/*number is an int*/) {
[retnArr addObject:[NSNumber numberWithInt:/*current number*/]];
}
}
return retnArr;
}
Stick your logic into this method. Once you've found a number you want to return, stick it into the array using the [retnArr addObject:[NSNumber numberWithInt:]]; method I put up there.
Once you've returned the array, access the numbers like this:
[[arrReturnedFromMethod objectAtIndex:someIndex] intValue];
Optionally, you might want to throw them into the NSNumber object as different types.
You can also use:
[NSNumber numberWithDouble:]
[NSNumber numberWithLongLong:]
And there are matching getters (doubleValue,longLongValue) to extract the number. There are lots of other methods for NSNumber, but these seem the most likely you'd want to be using.
Not to get confused with the NSString sizeWithFont method that returns a CGSize, what I'm looking for is a method that returns an NSString constrained to a certain CGSize. The reason I want to do this is so that when drawing text with Core Text, I can get append an ellipses (...) to the end of the string. I know NSString's drawInRect method does this for me, but I'm using Core Text, and kCTLineBreakByTruncatingTail truncates the end of each line rather than the end of the string.
There's this method that I found that truncates a string to a certain width, and it's not that hard to change it to make it work for a CGSize, but the algorithm is unbelievably slow for long strings, and is practically unusable. (It took over 10 seconds to truncate a long string). There has to be a more "computer science"/mathematical algorithm way to do this faster. Anyone daring enough to try to come up with a faster implementation?
Edit: I've managed to make this in to a binary algorithm:
-(NSString*)getStringByTruncatingToSize:(CGSize)size string:(NSString*)string withFont:(UIFont*)font
{
int min = 0, max = string.length, mid;
while (min < max) {
mid = (min+max)/2;
NSString *currentString = [string substringWithRange:NSMakeRange(min, mid - min)];
CGSize currentSize = [currentString sizeWithFont:font constrainedToSize:CGSizeMake(size.width, MAXFLOAT)];
if (currentSize.height < size.height){
min = mid + 1;
} else if (currentSize.height > size.height) {
max = mid - 1;
} else {
break;
}
}
NSMutableString *finalString = [[string substringWithRange:NSMakeRange(0, min)] mutableCopy];
if(finalString.length < self.length)
[finalString replaceCharactersInRange:NSMakeRange(finalString.length - 3, 3) withString:#"..."];
return finalString;
}
The problem is that this sometimes cuts the string too short when it has room to spare. I think this is where that last condition comes in to play. How do I make sure it doesn't cut off too much?
Good news! There is a "computer science/mathematical way" to do this faster.
The example you link to does a linear search: it just chops one character at a time from the end of the string until it's short enough. So, the amount of time it takes will scale linearly with the length of the string, and with long strings it will be really slow, as you've discovered.
However, you can easily apply a binary search technique to the string. Instead of starting at the end and dropping off one character at a time, you start in the middle:
THIS IS THE STRING THAT YOU WANT TO TRUNCATE
^
You compute the width of "THIS IS THE STRING THAT". If it is too wide, you move your test point to the midpoint of the space on the left. Like this:
THIS IS THE STRING THAT YOU WANT TO TRUNCATE
^ |
On the other hand, if it isn't wide enough, you move the test point to the midpoint of the other half:
THIS IS THE STRING THAT YOU WANT TO TRUNCATE
| ^
You repeat this until you find the point that is just under your width limit. Because you're dividing your search area in half each time, you'll never need to compute the width more than log2 N times (where N is the length of the string) which doesn't grow very fast, even for very long strings.
To put it another way, if you double the length of your input string, that's only one additional width computation.
Starting with Wikipedia's binary search sample, here's an example. Note that since we're not looking for an exact match (you want largest that will fit) the logic is slightly different.
int binary_search(NSString *A, float max_width, int imin, int imax)
{
// continue searching while [imin,imax] is not empty
while (imax >= imin)
{
/* calculate the midpoint for roughly equal partition */
int imid = (imin + imax) / 2;
// determine which subarray to search
float width = ComputeWidthOfString([A substringToIndex:imid]);
if (width < max_width)
// change min index to search upper subarray
imin = imid + 1;
else if (width > max_width )
// change max index to search lower subarray
imax = imid - 1;
else
// exact match found at index imid
return imid;
}
// Normally, this is the "not found" case, but we're just looking for
// the best fit, so we return something here.
return imin;
}
You need to do some math or testing to figure out what's the right index at the bottom, but it's definitely imin or imax, plus or minus one.