As the title.
I tested NSScanner, but it passed some strange strings. (ex :123aaa).
Is there any way to convert string<->number strictly?
You can easily roll your own. Test whether the entire string was scanned, or whether there are additional characters.
NSScanner *scanner = [NSScanner localizedScannerWithString:str];
int i;
if (![scanner scanInt:&i] || [scanner scanLocation] < [str length]) {
// str contains additional characters
...
} else {
// str contains only an int
...
}
NSScanner isn't that high-level. You'll have to validate the string yourself.
One way would be to scan characters up to the set of digits, assert that that failed, then scan the digits, then scan to the end and assert that that failed.
Related
I'm looking for a way to replace emoji characters with their description in a Swift string.
Example:
Input "This is my string 😄"
I'd like to replace the 😄 to get:
Output "This is my string {SMILING FACE WITH OPEN MOUTH AND SMILING EYES}"
To date I'm using this code modified from the original code of this answer by MartinR, but it works only if I deal with a single character.
let myCharacter : Character = "😄"
let cfstr = NSMutableString(string: String(myCharacter)) as CFMutableString
var range = CFRangeMake(0, CFStringGetLength(cfstr))
CFStringTransform(cfstr, &range, kCFStringTransformToUnicodeName, Bool(0))
var newStr = "\(cfstr)"
// removing "\N" from the result: \N{SMILING FACE WITH OPEN MOUTH AND SMILING EYES}
newStr = newStr.stringByReplacingOccurrencesOfString("\\N", withString:"")
print("\(newStr)") // {SMILING FACE WITH OPEN MOUTH AND SMILING EYES}
How can I achieve this?
Simply do not use a Character in the first place but use a String as input:
let cfstr = NSMutableString(string: "This 😄 is my string 😄") as CFMutableString
that will finally output
This {SMILING FACE WITH OPEN MOUTH AND SMILING EYES} is my string {SMILING FACE WITH OPEN MOUTH AND SMILING EYES}
Put together:
func transformUnicode(input : String) -> String {
let cfstr = NSMutableString(string: input) as CFMutableString
var range = CFRangeMake(0, CFStringGetLength(cfstr))
CFStringTransform(cfstr, &range, kCFStringTransformToUnicodeName, Bool(0))
let newStr = "\(cfstr)"
return newStr.stringByReplacingOccurrencesOfString("\\N", withString:"")
}
transformUnicode("This 😄 is my string 😄")
Here is a complete implementation.
It avoids to convert to description also the non-emoji characters (e.g. it avoids to convert “ to {LEFT DOUBLE QUOTATION MARK}). To accomplish this, it uses an extension based on this answer by Arnold that returns true or false whether a string contains an emoji.
The other part of the code is based on this answer by MartinR and the answer and comments to this answer by luk2302.
var str = "Hello World 😄 …" // our string (with an emoji and a horizontal ellipsis)
let newStr = str.characters.reduce("") { // loop through str individual characters
var item = "\($1)" // string with the current char
let isEmoji = item.containsEmoji // true or false
if isEmoji {
item = item.stringByApplyingTransform(String(kCFStringTransformToUnicodeName), reverse: false)!
}
return $0 + item
}.stringByReplacingOccurrencesOfString("\\N", withString:"") // strips "\N"
extension String {
var containsEmoji: Bool {
for scalar in unicodeScalars {
switch scalar.value {
case 0x1F600...0x1F64F, // Emoticons
0x1F300...0x1F5FF, // Misc Symbols and Pictographs
0x1F680...0x1F6FF, // Transport and Map
0x2600...0x26FF, // Misc symbols
0x2700...0x27BF, // Dingbats
0xFE00...0xFE0F, // Variation Selectors
0x1F900...0x1F9FF: // Various (e.g. 🤖)
return true
default:
continue
}
}
return false
}
}
print (newStr) // Hello World {SMILING FACE WITH OPEN MOUTH AND SMILING EYES} …
Please note that some emoji could not be included in the ranges of this code, so you should check if all the emoji are converted at the time you will implement the code.
I have a problem in regards of extracting signed int from string in c++.
Assuming that i have a string of images1234, how can i extract the 1234 from the string without knowing the position of the last non numeric character in C++.
FYI, i have try stringstream as well as lexical_cast as suggested by others through the post but stringstream returns 0 while lexical_cast stopped working.
int main()
{
string virtuallive("Images1234");
//stringstream output(virtuallive.c_str());
//int i = stoi(virtuallive);
//stringstream output(virtuallive);
int i;
i = boost::lexical_cast<int>(virtuallive.c_str());
//output >> i;
cout << i << endl;
return 0;
}
How can i extract the 1234 from the string without knowing the position of the last non numeric character in C++?
You can't. But the position is not hard to find:
auto last_non_numeric = input.find_last_not_of("1234567890");
char* endp = &input[0];
if (last_non_numeric != std::string::npos)
endp += last_non_numeric + 1;
if (*endp) { /* FAILURE, no number on the end */ }
auto i = strtol(endp, &endp, 10);
if (*endp) {/* weird FAILURE, maybe the number was really HUGE and couldn't convert */}
Another possibility would be to put the string into a stringstream, then read the number from the stream (after imbuing the stream with a locale that classifies everything except digits as white space).
// First the desired facet:
struct digits_only: std::ctype<char> {
digits_only(): std::ctype<char>(get_table()) {}
static std::ctype_base::mask const* get_table() {
// everything is white-space:
static std::vector<std::ctype_base::mask>
rc(std::ctype<char>::table_size,std::ctype_base::space);
// except digits, which are digits
std::fill(&rc['0'], &rc['9'], std::ctype_base::digit);
// and '.', which we'll call punctuation:
rc['.'] = std::ctype_base::punct;
return &rc[0];
}
};
Then the code to read the data:
std::istringstream virtuallive("Images1234");
virtuallive.imbue(locale(locale(), new digits_only);
int number;
// Since we classify the letters as white space, the stream will ignore them.
// We can just read the number as if nothing else were there:
virtuallive >> number;
This technique is useful primarily when the stream contains a substantial amount of data, and you want all the data in that stream to be interpreted in the same way (e.g., only read numbers, regardless of what else it might contain).
I have an array with a bunch of strings and I want to check if a certain string is contained in the array. If I use the containsObject: message on the array, I'm getting correct results. Do all NSString objects with the same string point to the same object? Or why is the containsObject: working?
NSArray *stringArray = [NSArray arrayWithObjects:#"1",#"2",#"3",anotherStringValue, nil];
if([stringArray containsObject:#"2"]){
//DO SOMETHING
}
Yes, hard-coded NSStrings (string literals) (that is any #"..." in your source code) are turned into strings that exist indefinitely while your process is running.
However NSArray's containsObject: methods calls isEqual: on its objects, hence even a dynamically created string such as [NSString stringWithFormat:#"%d", 2] would return YES in your sample snippet.
This is because NSString's isEqual: (or more precisely its isEqualToString:) method is implemented to be content aware (vs. comparing pointer identities) and thus returns YES for any pair of strings containing the very same sequence of characters (at time of comparison), no matter how and when they were created.
To check for equal (pointer-)identity you'd have to enumerate your array and compare via
NSString *yourString = #"foo";
BOOL identicalStringFound = NO;
for (NSString *someString in stringArray) {
if (someString == yourString) {
identicalStringFound = YES;
break;
}
}
(which you most likely wouldn't want, though).
Or in a more convenient fashion:
BOOL identicalStringFound = [stringArray indexOfObjectIdenticalTo:someString] != NSNotFound;
(you most likely wouldn't want this one either).
Summing up:
So the reason you're getting a positive reply from containsObject: is NOT because literal strings share the same constant instance, BUT because containsObject: by convention calls isEqual:, which is content aware.
You might want to read the (short) documentation for isEqual: from the NSObject protocol.
containsObject: performs a value check, not a pointer check. It uses the isEqual: method defined by NSObject and overridden by other objects for testing. Therefore, if two strings contain the same sequence of characters, they will be considered the same.
The distinction between pointer testing and value testing is very important in some cases. Constant strings defined in source code are combined by the compiler so that they are the same object. However, strings created dynamically are not the same object. Here is an example program which will demonstrate this:
int main(int argc, char **argv) {
NSAutoreleasePool *p = [NSAutoreleasePool new];
NSString *constantString = #"1";
NSString *constantString2 = #"1";
NSString *dynamicString = [NSString stringWithFormat:#"%i",1];
NSArray *theArray = [NSArray arrayWithObject:constantString];
if(constantString == constantString2) NSLog(#"constantString == constantString2");
else NSLog(#"constantString != constantString2");
if(constantString == dynamicString) NSLog(#"constantString == dynamicString");
else NSLog(#"constantString != dynamicString");
if([constantString isEqual:dynamicString]) NSLog(#"[constantString isEqual:dynamicString] == YES");
else NSLog(#"[constantString isEqual:dynamicString] == NO");
NSLog(#"theArray contains:\n\tconstantString: %i\n\tconstantString2: %i\n\tdynamicString: %i",
[theArray containsObject:constantString],
[theArray containsObject:constantString2],
[theArray containsObject:dynamicString]);
}
The output of this program is:
2011-04-27 17:10:54.686 a.out[41699:903] constantString == constantString2
2011-04-27 17:10:54.705 a.out[41699:903] constantString != dynamicString
2011-04-27 17:10:54.706 a.out[41699:903] [constantString isEqual:dynamicString] == YES
2011-04-27 17:10:54.706 a.out[41699:903] theArray contains:
constantString: 1
constantString2: 1
dynamicString: 1
You can use containsObject to findout if certain string is exist,
NSArray *stringArray = [NSArray arrayWithObjects:#"1",#"2",#"3",anotherStringValue, nil];
if ( [stringArray containsObject: stringToFind] ) {
// if found
} else {
// if not found
}
I am trying to convert an decimal number to it's character equivalent. For example:
int j = 65 // The character equivalent would be 'A'.
Sorry, forgot to specify the language. I thought I did. I am using the Cocoa/Object-C. It is really frustrating. I have tried the following but it is still not converting correctly.
char_num1 = [working_text characterAtIndex:i]; // value = 65
char_num2 = [working_text characterAtIndex:i+1]; // value = 75
char_num3 = char_num1 + char_num2; // value = 140
char_str1 = [NSString stringWithFormat:#"%c",char_num3]; // mapped value = 229
char_str2 = [char_str2 stringByAppendingString:char_str1];
When char_num1 and char_num2 are added, I get the new ascii decimal value. However, when I try to convert the new decimal value to a character, I do not get the character that is mapped to char_num3.
Convert a character to a number in C:
int j = 'A';
Convert a number to a character in C:
char ch = 65;
Convert a character to a number in python:
j = ord('A')
Convert a number to a character in Python:
ch = chr(65)
Most languages have a 'char' function, so it would be Char(j)
I'm not sure what language you're asking about. In Java, this works:
int a = 'a';
It's quite often done with "chr" or "char", but some indication of the language / platform would be useful :-)
string k = Chr(j);
Say I have an NSString (or NSMutableString) containing:
I said "Hello, world!".
He said "My name's not World."
What's the best way to turn that into:
I said \"Hello, world!\".\nHe said \"My name\'s not World.\"
Do I have to manually use -replaceOccurrencesOfString:withString: over and over to escape characters, or is there an easier way? These strings may contain characters from other alphabets/languages.
How is this done in other languages with other string classes?
stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding
I don't think there is any built-in method to "escape" a particular set of characters.
If the characters you wish to escape is well-defined, I'd probably stick with the simple solution you proposed, replacing the instances of the characters crudely.
Be warned that if your source string already has escaped characters in it, then you'll probably want to avoid "double-escaping" them. One way of achieving this would be to go through and "unescape" any escaped character strings in the string before then escaping them all again.
If you need to support a variable set of escaped characters, take a look at the NSScanner methods "scanUpToCharactersFromSet:intoString:" and "scanCharactersFromSet:intoString:". You could use these methods on NSScanner to cruise through a string, copying the parts from the "scanUpTo" section into a mutable string unchanged, and copying the parts from a particular character set only after escaping them.
This will escape double quotes in NSString:
NSString *escaped = [originalString stringByReplacingOccurrencesOfString:#"\"" withString:#"\\\""];
So you need to be careful and also escape the escape character...
I think in cases like these, it's useful to operate on a character at a time, either in UniChars or UTF8 bytes. If you're using UTF-8, then vis(3) will do most of the work for you (see below). Can I ask why you want to escape a single-quote within a double-quoted string? How are you planning to handle multi-byte characters? In the example below, I'm using UTF-8, encoding 8-bit characters using C-Style octal escapes. This can also be undone by unvis(3).
#import <Foundation/Foundation.h>
#import <vis.h>
#interface NSString (Escaping)
- (NSString *)stringByEscapingMetacharacters;
#end
#implementation NSString (Escaping)
- (NSString *)stringByEscapingMetacharacters
{
const char *UTF8Input = [self UTF8String];
char *UTF8Output = [[NSMutableData dataWithLength:strlen(UTF8Input) * 4 + 1 /* Worst case */] mutableBytes];
char ch, *och = UTF8Output;
while ((ch = *UTF8Input++))
if (ch == '\'' || ch == '\'' || ch == '\\' || ch == '"')
{
*och++ = '\\';
*och++ = ch;
}
else if (isascii(ch))
och = vis(och, ch, VIS_NL | VIS_TAB | VIS_CSTYLE, *UTF8Input);
else
och+= sprintf(och, "\\%03hho", ch);
return [NSString stringWithUTF8String:UTF8Output];
}
#end
int
main(int argc, const char *argv[])
{
NSAutoreleasePool *pool = [NSAutoreleasePool new];
NSLog(#"%#", [#"I said \"Hello, world!\".\nHe said \"My name's not World.\"" stringByEscapingMetacharacters]);
[pool drain];
return 0;
}
This is a snippet I have used in the past that works quite well:
- (NSString *)escapeString:(NSString *)aString
{
NSMutableString *returnString = [[NSMutableString alloc] init];
for(int i = 0; i < [aString length]; i++) {
unichar c = [aString characterAtIndex:i];
// if char needs to be escaped
if((('\\' == c) || ('\'' == c)) || ('"' == c)) {
[returnString appendFormat:#"\\%c", c];
} else {
[returnString appendFormat:#"%c", c];
}
}
return [returnString autorelease];
}
Do this:
NSString * encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(
NULL,
(CFStringRef)unencodedString,
NULL,
(CFStringRef)#"!*'();:#&=+$,/?%#[]",
kCFStringEncodingUTF8 );
Reference: http://simonwoodside.com/weblog/2009/4/22/how_to_really_url_encode/
You might even want to look into using a regex library (there are a lot of options available, RegexKit is a popular choice). It shouldn't be too hard to find a pre-written regex to escape strings that handles special cases like existing escaped characters.