Anybody has some code in objective-c to convert a NSInteger or NSString to binary string?
example:
56 -> 111000
There are some code in stackoverflow that try do this, but it doesn´t work.
Thanks
Not sure which examples on SO didn't work for you, but Adam Rosenfield's answer here seems to work. I've updated it to remove a compiler warning:
// Original author Adam Rosenfield... SO Question 655792
NSInteger theNumber = 56;
NSMutableString *str = [NSMutableString string];
for(NSInteger numberCopy = theNumber; numberCopy > 0; numberCopy >>= 1)
{
// Prepend "0" or "1", depending on the bit
[str insertString:((numberCopy & 1) ? #"1" : #"0") atIndex:0];
}
NSLog(#"Binary version: %#", str);
Tooting my own horn a bit here...
I've written a math framework called CHMath that deals with arbitrarily large integers. One of the things it does is allows the user to create a CHNumber from a string, and get its binary representation as a string. For example:
CHNumber * t = [CHNumber numberWithString:#"56"];
NSString * binaryT = [t binaryStringValue];
NSLog(#"Binary value of %#: %#", t, binaryT);
Logs:
2009-12-15 10:36:10.595 otest-x86_64[21918:903] Binary value of 56: 0111000
The framework is freely available on its Github repository.
Related
I have an output string in this format .
I need to format the string such that i can display the URL separately and my Content, the description separately. Is there any functions , so i can format them easily ?
The code :
NSLog(#"Description %#", string);
The OUTPUT String:
2013-07-28 11:13:59.083 RSSreader[4915:c07] Description
http://www.apple.com/pr/library/2013/07/23Apple-Reports-Third-Quarter-Results.html?sr=hotnews.rss
Apple today announced financial results for its fiscal 2013 third quarter ended
June 29, 2013. The Company posted quarterly revenue of $35.3 billion and quarterly
net profit of $6.9 billion, or $7.47 per diluted share.
Apple sold 31.2 million iPhones, which set a June quarter record.
You should extract URL from string, then display it in formatted way.
A simple way to extracting URL is regular expressions (RegEX).
After extracting URL you can replace it with nothing:
str = [str stringByReplacingOccurrencesOfString:extractedURL
withString:#""];
You can use this :
https://stackoverflow.com/a/9587987/305135
If description string separated by line break (\n), you can do this:
NSArray *items = [yourString componentsSeparatedByString:#"\n"];
Regex is a good idea.
But there is a default way of detecting URLs within a String in Objective C, NSDataDetector.
NSDataDetector internally uses Regex.
NSString *aString = #"YOUR STRING WITH URLs GOES HERE"
NSDataDetector *aDetector = [NSDataDetector dataDetectorWithTypes:NSTextCheckingTypeLink error:nil];
NSArray *theMatches = [aDetector matchesInString:aString options:0 range:NSMakeRange(0, [aString length])];
for (int anIndex = 0; anIndex < [theMatches count]; anIndex++) {
// If needed, Save this URL for Future Use.
NSString *anURLString = [[[theMatches objectAtIndex:anIndex] URL] absoluteString];
// Replace the Url with Empty String
aTitle = [aTitle stringByReplacingOccurrencesOfString:anURLString withString:#""];
}
I want to solve a conditional equation in iOS:
The equation I get from database is in NSString format, for example:
if((height > 0), (weight+ 2 ), ( weight-1 ) )
As per our understanding, if I parse the above string and separateheight>0condition, it will be in the NSString format. But to evaluate it how do I convert the string to a conditional statement?
Once the conditional statement is obtained the equation can be solved by converting it to a ternary equation as follows:
Bool status;
NSString *condition=#” height>0”;
If(condition) //But condition is treated as a string and not as a conditional statement.
{
status=True;
}
else
{
status=False;
}
Return status ? weight+ 2 : weight-1;`
Also the equations can dynamically change, so they cannot be hard coded. In short how do I solve this equation which I get as a NSString.
Thank you for your patience!
DDMathParser author here...
To expand on Jonathan's answer, here's how you could do it entirely in DDMathParser. However, to parse the string as-is, you'll need to do two things.
First, you'll need to create an if function:
DDMathEvaluator *evaluator = [DDMathEvaluator sharedMathEvaluator];
[evaluator registerFunction:^DDExpression *(NSArray *args, NSDictionary *vars, DDMathEvaluator *eval, NSError *__autoreleasing *error) {
if ([args count] == 3) {
DDExpression *condition = [args objectAtIndex:0];
DDExpression *resultExpression = nil;
NSNumber *conditionValue = [condition evaluateWithSubstitutions:vars evaluator:eval error:error];
if ([conditionValue boolValue] == YES) {
resultExpression = [args objectAtIndex:1];
} else {
resultExpression = [args objectAtIndex:2];
}
NSNumber *result = [resultExpression evaluateWithSubstitutions:vars evaluator:eval error:error];
return [DDExpression numberExpressionWithNumber:result];
}
return nil;
} forName:#"if"];
This creates the if() function, which takes three parameters. Depending on how the first parameter evaluates, it either evaluates to the result of the second or third parameter.
The other thing you'll need to do is tell the evaluator what height and weight mean. Since they don't start with a $ character, they get interpreted as functions, and not variables. If they started with a $, then it would be as simple as evaluating it like this:
NSString *expression = #"if(($height > 0), ($weight+ 2 ), ( $weight-1 ) )";
NSDictionary *variables = #{#"height" : #42, #"weight" : #13};
NSNumber *result = [expression evaluateWithSubstitutions:variables evaluator:evaluator error:nil];
However, since they don't start with a $, they're functions, which means you need to tell the evaluator what the functions evaluate to. You could do this by creating functions for both height and weight, just like you did for if:
[evaluator registerFunction:^DDExpression *(NSArray *args, NSDictionary *vars, DDMathEvaluator *eval, NSError **error) {
return [DDExpression numberExpressionWithNumber:#42];
} forName:#"height"];
Alternatively, you could make it a bit more dynamic and use the functionResolver block of DDMathEvaluator, which is a block that returns a block (woooooo) and would look like this:
NSDictionary *values = #{#"height": #42, #"weight": #13};
[evaluator setFunctionResolver:^DDMathFunction(NSString *name) {
DDMathFunction f = ^(NSArray *args, NSDictionary *vars, DDMathEvaluator *eval, NSError **error) {
NSNumber *n = [values objectForKey:name];
if (!n) { n = #0; }
return [DDExpression numberExpressionWithNumber:n];
};
return f;
}];
With those two pieces in place (registering if and providing the values of height and weight), you can do:
NSString *expression = #"if((height > 0), (weight+ 2 ), ( weight-1 ) )";
NSNumber *result = [expression evaluateWithSubstitutions:nil evaluator:evaluator error:nil];
... and get back the proper result of #15.
(I have plans to make DDMathParser allow unknown functions to fall back to provided variable values, but I haven't quite finished it yet)
you will have to write your own interpreter or find one that supports this kind of expressions.
The first part (the condition) can be evaluated by NSPredicate. For the second part (the calculation) you will need some math expression evaluation. Try this out https://github.com/davedelong/DDMathParser. Maybe you can do both with DDMathParser but i am not sure.
NSString *string = [myString stringByReplacingOccurrencesOfString:#"<wow>" withString:someString];
I have this code. Now suppose my app's user enters two different strings I want to replace with two different other strings, how do I achieve that? I don't care if it uses private APIs, i'm developing for the jailbroken platform. My user is going to either enter or or . I want to replace any occurrences of those strings with their respective to-be-replaced-with strings :)
Thanks in advance :P
Both dasblinkenlight’s and Matthias’s answers will work, but they both result in the creation of a couple of intermediate NSStrings; that’s not really a problem if you’re not doing this operation often, but a better approach would look like this.
NSMutableString *myStringMut = [[myString mutableCopy] autorelease];
[myStringMut replaceOccurrencesOfString:#"a" withString:somethingElse];
[myStringMut replaceOccurrencesOfString:#"b" withString:somethingElseElse];
// etc.
You can then use myStringMut as you would’ve used myString, since NSMutableString is an NSString subclass.
The simplest solution is running stringByReplacingOccurrencesOfString twice:
NSString *string = [[myString
stringByReplacingOccurrencesOfString:#"<wow>" withString:someString1]
stringByReplacingOccurrencesOfString:#"<boo>" withString:someString2];
I would just run the string replacing method again
NSString *string = [myString stringByReplacingOccurrencesOfString:#"foo" withString:#"String 1"];
string = [string stringByReplacingOccurrencesOfString:#"bar" withString:#"String 2"];
This works well for me in Swift 3.1
let str = "hi hello hey"
var replacedStr = (str as NSString).replacingOccurrences(of: "hi", with: "Hi")
replacedStr = (replacedStr as NSString).replacingOccurrences(of: "hello", with: "Hello")
replacedStr = (replacedStr as NSString).replacingOccurrences(of: "hey", with: "Hey")
print(replacedStr) // Hi Hello Hey
I have a string that is being generate from a formula, however I only want to use the string as long as all of its characters are numeric, if not that I want to do something different for instance display an error message.
I have been having a look round but am finding it hard to find anything that works along the lines of what I am wanting to do. I have looked at NSScanner but I am not sure if its checking the whole string and then I am not actually sure how to check if these characters are numeric
- (void)isNumeric:(NSString *)code{
NSScanner *ns = [NSScanner scannerWithString:code];
if ( [ns scanFloat:NULL] ) //what can I use instead of NULL?
{
NSLog(#"INSIDE IF");
}
else {
NSLog(#"OUTSIDE IF");
}
}
So after a few more hours searching I have stumbled across an implementation that dose exactly what I am looking for.
so if you are looking to check if their are any alphanumeric characters in your NSString this works here
-(bool) isNumeric:(NSString*) hexText
{
NSNumberFormatter* numberFormatter = [[[NSNumberFormatter alloc] init] autorelease];
NSNumber* number = [numberFormatter numberFromString:hexText];
if (number != nil) {
NSLog(#"%# is numeric", hexText);
//do some stuff here
return true;
}
NSLog(#"%# is not numeric", hexText);
//or do some more stuff here
return false;
}
hope this helps.
Something like this would work:
#interface NSString (usefull_stuff)
- (BOOL) isAllDigits;
#end
#implementation NSString (usefull_stuff)
- (BOOL) isAllDigits
{
NSCharacterSet* nonNumbers = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
NSRange r = [self rangeOfCharacterFromSet: nonNumbers];
return r.location == NSNotFound && self.length > 0;
}
#end
then just use it like this:
NSString* hasOtherStuff = #"234 other stuff";
NSString* digitsOnly = #"123345999996665003030303030";
BOOL b1 = [hasOtherStuff isAllDigits];
BOOL b2 = [digitsOnly isAllDigits];
You don't have to wrap the functionality in a private category extension like this, but it sure makes it easy to reuse..
I like this solution better than the others since it wont ever overflow some int/float that is being scanned via NSScanner - the number of digits can be pretty much any length.
Consider NSString integerValue - it returns an NSInteger. However, it will accept some strings that are not entirely numeric and does not provide a mechanism to determine strings which are not numeric at all. This may or may not be acceptable.
For instance, " 13 " -> 13, "42foo" -> 42 and "helloworld" -> 0.
Happy coding.
Now, since the above was sort of a tangent to the question, see determine if string is numeric. Code taken from link, with comments added:
BOOL isNumeric(NSString *s)
{
NSScanner *sc = [NSScanner scannerWithString: s];
// We can pass NULL because we don't actually need the value to test
// for if the string is numeric. This is allowable.
if ( [sc scanFloat:NULL] )
{
// Ensure nothing left in scanner so that "42foo" is not accepted.
// ("42" would be consumed by scanFloat above leaving "foo".)
return [sc isAtEnd];
}
// Couldn't even scan a float :(
return NO;
}
The above works with just scanFloat -- e.g. no scanInt -- because the range of a float is much larger than that of an integer (even a 64-bit integer).
This function checks for "totally numeric" and will accept "42" and "0.13E2" but reject " 13 ", "42foo" and "helloworld".
It's very simple.
+ (BOOL)isStringNumeric:(NSString *)text
{
NSCharacterSet *alphaNums = [NSCharacterSet decimalDigitCharacterSet];
NSCharacterSet *inStringSet = [NSCharacterSet characterSetWithCharactersInString:text];
return [alphaNums isSupersetOfSet:inStringSet];
}
Like this:
- (void)isNumeric:(NSString *)code{
NSScanner *ns = [NSScanner scannerWithString:code];
float the_value;
if ( [ns scanFloat:&the_value] )
{
NSLog(#"INSIDE IF");
// do something with `the_value` if you like
}
else {
NSLog(#"OUTSIDE IF");
}
}
Faced same problem in Swift.
In Swift you should use this code, according TomSwift's answer:
func isAllDigits(str: String) -> Bool {
let nonNumbers = NSCharacterSet.decimalDigitCharacterSet()
if let range = str.rangeOfCharacterFromSet(nonNumbers) {
return true
}
else {
return false
}
}
P.S. Also you can use other NSCharacterSets or their combinations to check your string!
For simple numbers like "12234" or "231231.23123" the answer can be simple.
There is a transformation law for int numbers: when string with integer transforms to int (or long) number and then, again, transforms it back to another string these strings will be equal.
In Objective C it will looks like:
NSString *numStr=#"1234",*num2Str=nil;
num2Str=[NSString stringWithFormat:#"%lld",numStr.longlongValue];
if([numStr isEqualToString: num2Str]) NSLog(#"numStr is an integer number!");
By using this transformation law we can create solution
to detect double or long numbers:
NSString *numStr=#"12134.343"
NSArray *numList=[numStr componentsSeparatedByString:#"."];
if([[NSString stringWithFormat:#"%lld", numStr.longLongValue] isEqualToString:numStr]) NSLog(#"numStr is an integer number");
else
if( numList.count==2 &&
[[NSString stringWithFormat:#"%lld",((NSString*)numList[0]).longLongValue] isEqualToString:(NSString*)numList[0]] &&
[[NSString stringWithFormat:#"%lld",((NSString*)numList[1]).longLongValue] isEqualToString:(NSString*)numList[1]] )
NSLog(#"numStr is a double number");
else
NSLog(#"numStr is not a number");
I did not copy the code above from my work code so can be some mistakes, but I think the main point is clear.
Of course this solution doesn't work with numbers like "1E100", as well it doesn't take in account size of integer and fractional part. By using the law described above you can do whatever number detection you need.
C.Johns' answer is wrong. If you use a formatter, you risk apple changing their codebase at some point and having the formatter spit out a partial result. Tom's answer is wrong too. If you use the rangeOfCharacterFromSet method and check for NSNotFound, it'll register a true if the string contains even one number. Similarly, other answers in this thread suggest using the Integer value method. That is also wrong because it will register a true if even one integer is present in the string. The OP asked for an answer that ensures the entire string is numerical. Try this:
NSCharacterSet *searchSet = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
Tom was right about this part. That step gives you the non-numerical string characters. But then we do this:
NSString *trimmedString = [string stringByTrimmingCharactersInSet:searchSet];
return (string.length == trimmedString.length);
Tom's inverted character set can TRIM a string. So we can use that trim method to test if any non numerals exist in the string by comparing their lengths.
I would like to get the percent encoded string for these specific letters, how to do that in objective-c?
Reserved characters after percent-encoding
! * ' ( ) ; : # & = + $ , / ? # [ ]
%21 %2A %27 %28 %29 %3B %3A %40 %26 %3D %2B %24 %2C %2F %3F %23 %5B %5D
Percent-encoding wiki
Please test with this string and see if it do work:
myURL = #"someurl/somecontent"
I would like the string to look like:
myEncodedURL = #"someurl%2Fsomecontent"
I tried with the stringByAddingPercentEscapesUsingEncoding: NSASCIIStringEncoding already but it does not work, the result is still the same as the original string. Please advice.
I've found that both stringByAddingPercentEscapesUsingEncoding: and CFURLCreateStringByAddingPercentEscapes() are inadequate. The NSString method misses quite a few characters, and the CF function only lets you say which (specific) characters you want to escape. The proper specification is to escape all characters except a small set.
To fix this, I created an NSString category method to properly encode a string. It will percent encoding everything EXCEPT [a-zA-Z0-9.-_~] and will also encode spaces as + (according to this specification). It will also properly handle encoding unicode characters.
- (NSString *) URLEncodedString_ch {
NSMutableString * output = [NSMutableString string];
const unsigned char * source = (const unsigned char *)[self UTF8String];
int sourceLen = strlen((const char *)source);
for (int i = 0; i < sourceLen; ++i) {
const unsigned char thisChar = source[i];
if (thisChar == ' '){
[output appendString:#"+"];
} else if (thisChar == '.' || thisChar == '-' || thisChar == '_' || thisChar == '~' ||
(thisChar >= 'a' && thisChar <= 'z') ||
(thisChar >= 'A' && thisChar <= 'Z') ||
(thisChar >= '0' && thisChar <= '9')) {
[output appendFormat:#"%c", thisChar];
} else {
[output appendFormat:#"%%%02X", thisChar];
}
}
return output;
}
The iOS 7 SDK now has a better alternative tostringByAddingPercentEscapesUsingEncoding that does let you specify that you want all characters escaped except certain allowed ones. It works well if you are building up the URL in parts:
NSString * unescapedQuery = [[NSString alloc] initWithFormat:#"?myparam=%d", numericParamValue];
NSString * escapedQuery = [unescapedQuery stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]];
NSString * urlString = [[NSString alloc] initWithFormat:#"http://ExampleOnly.com/path.ext%#", escapedQuery];
Although it's less often that the other parts of the URL will be variables, there are constants in the NSURLUtilities category for those as well:
[NSCharacterSet URLHostAllowedCharacterSet]
[NSCharacterSet URLUserAllowedCharacterSet]
[NSCharacterSet URLPasswordAllowedCharacterSet]
[NSCharacterSet URLPathAllowedCharacterSet]
[NSCharacterSet URLFragmentAllowedCharacterSet]
[NSCharacterSet URLQueryAllowedCharacterSet] includes all of the characters allowed in the query part of the URL (the part starting with the ? and before the # for a fragment, if any) including the ? and the & or = characters, which are used to delimit the parameter names and values. For query parameters with alphanumeric values, any of those characters might be included in the values of the variables used to build the query string. In that case, each part of the query string needs to be escaped, which takes just a bit more work:
NSMutableCharacterSet * URLQueryPartAllowedCharacterSet; // possibly defined in class extension ...
// ... and built in init or on first use
URLQueryPartAllowedCharacterSet = [[NSCharacterSet URLQueryAllowedCharacterSet] mutableCopy];
[URLQueryPartAllowedCharacterSet removeCharactersInString:#"&+=?"]; // %26, %3D, %3F
// then escape variables in the URL, such as values in the query and any fragment:
NSString * escapedValue = [anUnescapedValue stringByAddingPercentEncodingWithAllowedCharacters:URLQueryPartAllowedCharacterSet];
NSString * escapedFrag = [anUnescapedFrag stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLFragmentAllowedCharacterSet]];
NSString * urlString = [[NSString alloc] initWithFormat:#"http://ExampleOnly.com/path.ext?myparam=%##%#", escapedValue, escapedFrag];
NSURL * url = [[NSURL alloc] initWithString:urlString];
The unescapedValue could even be an entire URL, such as for a callback or redirect:
NSString * escapedCallbackParamValue = [anAlreadyEscapedCallbackURL stringByAddingPercentEncodingWithAllowedCharacters:URLQueryPartAllowedCharacterSet];
NSURL * callbackURL = [[NSURL alloc] initWithString:[[NSString alloc] initWithFormat:#"http://ExampleOnly.com/path.ext?callback=%#", escapedCallbackParamValue]];
Note: Don't use NSURL initWithScheme:(NSString *)scheme host:(NSString *)host path:(NSString *)path for a URL with a query string because it will add more percent escapes to the path.
NSString *encodedString = [myString stringByAddingPercentEscapesUsingEncoding:NSASCIIStringEncoding];
It won't replace your string inline; it'll return a new string. That's implied by the fact that the method starts with the word "string". It's a convenience method to instantiate a new instance of NSString based on the current NSString.
Note--that new string will be autorelease'd, so don't call release on it when you're done with it.
NSString's stringByAddingPercentEscapesUsingEncoding: looks like what you're after.
EDIT: Here's an example using CFURLCreateStringByAddingPercentEscapes instead. originalString can be either an NSString or a CFStringRef.
CFStringRef newString = CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, originalString, NULL, CFSTR("!*'();:#&=+#,/?#[]"), kCFStringEncodingUTF8);
Please note that this is untested. You should have a look at the documentation page to make sure you understand the memory allocation semantics for CFStringRef, the idea of toll-free bridging, and so on.
Also, I don't know (off the top of my head) which of the characters specified in the legalURLCharactersToBeEscaped argument would have been escaped anyway (due to being illegal in URLs). You may want to check this, although it's perhaps better just to be on the safe side and directly specify the characters you want escaped.
I'm making this answer a community wiki so that people with more knowledge about CoreFoundation can make improvements.
Following the RFC3986 standard, here is what I'm using for encoding URL components:
// https://tools.ietf.org/html/rfc3986#section-2.2
let rfc3986Reserved = NSCharacterSet(charactersInString: "!*'();:#&=+$,/?#[]")
let encoded = "email+with+plus#example.com".stringByAddingPercentEncodingWithAllowedCharacters(rfc3986Reserved.invertedSet)
Output: email%2Bwith%2Bplus%40example.com
If you are using ASI HttpRequest library in your objective-c program, which I cannot recommend highly enough, then you can use the "encodeURL" helper API on its ASIFormDataRequest object. Unfortunately, the API is not static so maybe worth creating an extension using its implementation in your project.
The code, copied straight from the ASIFormDataRequest.m for encodeURL implementation, is:
- (NSString*)encodeURL:(NSString *)string
{
NSString *newString = NSMakeCollectable([(NSString *)CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, (CFStringRef)string, NULL, CFSTR(":/?#[]#!$ &'()*+,;=\"<>%{}|\\^~`"), CFStringConvertNSStringEncodingToEncoding([self stringEncoding])) autorelease]);
if (newString) {
return newString;
}
return #"";
}
As you can see, it is essentially a wrapper around CFURLCreateStringByAddingPercentEscapes that takes care of all the characters that should be properly escaped.
Before I noticed Rob's answer, which appears to work well and is preferred as it's cleaner, I went ahead and ported Dave's answer to Swift. I'll leave it here in case anyone is interested:
public extension String {
// For performance, I've replaced the char constants with integers, as char constants don't work in Swift.
var URLEncodedValue: String {
let output = NSMutableString()
guard let source = self.cStringUsingEncoding(NSUTF8StringEncoding) else {
return self
}
let sourceLen = source.count
var i = 0
while i < sourceLen - 1 {
let thisChar = source[i]
if thisChar == 32 {
output.appendString("+")
} else if thisChar == 46 || thisChar == 45 || thisChar == 95 || thisChar == 126 ||
(thisChar >= 97 && thisChar <= 122) ||
(thisChar >= 65 && thisChar <= 90) ||
(thisChar >= 48 && thisChar <= 57) {
output.appendFormat("%c", thisChar)
} else {
output.appendFormat("%%%02X", thisChar)
}
i++
}
return output as String
}
}
In Swift4:
var str = "someurl/somecontent"
let percentEncodedString = str.addingPercentEncoding(withAllowedCharacters: .alphanumerics)