I need to convert some strings which have decimal content.
I want these contents in a string, in hex format...
How can I do it?
I tried to use NSScanner but this method seems to be a little bit huge to just make a dec-hex calculation?
Thanks for your answers!
If you know your string only contains a valid decimal number then the simplest way would be:
NSString *dec = #"254";
NSString *hex = [NSString stringWithFormat:#"0x%lX",
(unsigned long)[dec integerValue]];
NSLog(#"%#", hex);
Related
I have this line of code
// valueX is a long double (long double is a huge floating point)
NSString *value = [NSString stringWithFormat: #"%.10Lg", valueX];
This format specifier is specifying up to 10 decimal digits but I don't want to hard code this to 10.
I have this variable numberOfDigits that I want to be used to define the number of digits. For those itching to down vote this question, it is not so easy as it seems. I cannot substitute the 10 with %# because %.10Lg is a format specifier by itself.
OK, I can create a bunch of strings like #"%.5Lg", #"%.8Lg", #"%.9Lg"... and switch that, but I wonder if there is another way...
There is, if you read the manual pages for format specifiers. You can replace the precision with *, which means it will get taken from a parameter instead.
int numDigits = 10;
NSString *value = [NSString stringWithFormat:#"%.*Lg", numDigits, valueX];
I couldn't find this in the core foundation reference, but I know that this is written in the man 3 printf man page.
Dietrich's answer is the simplest and therefore best. Note that even if there wasn't a built-in way to specify the number of digits with a parameter you could still have done it by first building your format string and then using it:
- (NSString *) stringFromValue: (long double) value digits: (int) digits; {
//First create a format string. Use "%%" to escape the % escape char.
NSString *formatString =[NSString stringWithFormat: #"%%.%dLg", digits];
return [NSString stringWithFormat: formatString, value];
}
I've been working on a calculator and I wanted to implement conversions from decimal to octal and from decimal to hexadecimal. I'm new to Xcode and Objective C, but I've managed to get a conversion from decimal to octal, it just doesn't seem to work with hexadecimal.
Here's the code I've written to convert a double to octal:
double result = 0;
...
double decToOct = [self popOperand];
NSString *oct = [NSString stringWithFormat:#"%llo", (long long)decToOct];
result = [oct doubleValue];
Using the same scheme (obviously that includes changing #"%llo" with #"%llx") the conversion to hexadecimal works up to a certain point. It does numbers 0 through 9 just fine, but once it hits 10, it comes up as 0. To test, I also input 5395 and it displayed 1513, the desired result.
Because of this, I can only assume that for some reason my code does not want to input the actual letters of the hexadecimal values (e.g. 11 would convert to B but it shows up as 0) .
Any suggestions? Thanks in advance.
UPDATE:
In addition, I have also been using this to display the result:
double result = [self.brain performOperation:operation];
self.display.text = [NSString stringWithFormat:#"%g", result];
result, as listed from the top, is an argument which is eventually returned here, to self.brain performOperation:operation. This is supposed to handle the display of all operations, including: addition, multiplication, etc. but also octal and hexadecimal. Again, it works fine with octal, but not with hexadecimal.
Try this, May be it will help you. Please do let me know if i am wrong here:--->
NSString *decStr = #"11";
NSString *hexStr = [NSString stringWithFormat:#"%lX",
(unsigned long)[dec integerValue]];
NSLog(#"%#", hexStr);
If you know your string only contains a valid decimal number then the simplest way would be:
NSString *dec = #"254";
NSString *hex = [NSString stringWithFormat:#"0x%lX",
(unsigned long)[dec integerValue]];
NSLog(#"%#", hex);
I want to get a character from somewhere inside an NSString. I want the result to be an NSString.
This is the code I use to get a single character at index it:
[[s substringToIndex:i] substringToIndex:1]
Is there a better way to do it?
This will also retrieve a character at index i as an NSString, and you're only using an NSRange struct rather than an extra NSString.
NSString * newString = [s substringWithRange:NSMakeRange(i, 1)];
If you just want to get one character from an a NSString, you can try this.
- (unichar)characterAtIndex:(NSUInteger)index;
Used like so:
NSString *originalString = #"hello";
int index = 2;
NSString *theCharacter = [NSString stringWithFormat:#"%c", [originalString characterAtIndex:index-1]];
//returns "e".
Your suggestion only works for simple characters like ASCII. NSStrings store unicode and if your character is several unichars long then you could end up with gibberish. Use
- (NSRange)rangeOfComposedCharacterSequenceAtIndex:(NSUInteger)index;
if you want to determine how many unichars your character is. I use this to step through my strings to determine where the character borders occur.
Being fully unicode able is a bit of work but depends on what languages you use. I see a lot of asian text so most characters spill over from one space and so it's work that I need to do.
NSMutableString *myString=[NSMutableString stringWithFormat:#"Malayalam"];
NSMutableString *revString=#"";
for (int i=0; i<myString.length; i++) {
revString=[NSMutableString stringWithFormat:#"%c%#",[myString characterAtIndex:i],revString];
}
NSLog(#"%#",revString);
I want the user to input a string and then assign the input to an NSString. Right now my code looks like this:
NSString *word;
scanf("%s", &word);
The scanf function reads into a C string (actually an array of char), like this:
char word[40];
int nChars = scanf("%39s", word); // read up to 39 chars (leave room for NUL)
You can convert a char array into NSString like this:
NSString* word2 = [NSString stringWithBytes:word
length:nChars
encoding:NSUTF8StringEncoding];
However scanf only works with console (command line) programs. If you're trying to get input on a Mac or iOS device then scanf is not what you want to use to get user input.
scanf does not work with any object types. If you have a C string and want to create an NSString from it, use -[NSString initWithBytes:length:encoding:].
scanf does not work with NSString as scanf doesn’t work on objects. It works only on primitive datatypes such as:
int
float
BOOL
char
What to do?
Technically a string is made up of a sequence of individual characters. So to accept string input, you can read in the sequence of characters and convert it to a string.
use:
[NSString stringWithCString:cstring encoding:1];
Here is a working example:
NSLog(#"What is the first name?");
char cstring[40];
scanf("%s", cstring);
firstName = [NSString stringWithCString:cstring encoding:1];
Here’s an explanation of the above code, comment by comment:
You declare a variable called cstring to hold 40 characters.
You then tell scanf to expect a list of characters by using the %s format specifier.
Finally, you create an NSString object from the list of characters that were read in.
Run your project; if you enter a word and hit Enter, the program should print out the same word you typed. Just make sure the word is less than 40 characters; if you enter more, you might cause the program to crash — you are welcome to test that out yourself! :]
Taken from: RW.
This is how I'd do it:
char word [40];
scanf("%s",word);
NSString * userInput = [[NSString alloc] initWithCString: word encoding: NSUTF8StringEncoding];
yes, but sscanf does, and may be a good solution for complex NSString parsing.
Maybe this will work for you because it accepts string with spaces as well.
NSLog(#"Enter The Name Of State");
char name[20];
gets(name);
NSLog(#"%s",name);
Simple Solution is
char word[40];
scanf("%39s", word);
NSString* word2 = [NSString stringWithUTF8String:word];
The NSFileHandle class is an object-oriented wrapper for a file descriptor. For files, you can read, write, and seek within the file.
NSFileHandle *inputFile = [NSFileHandle fileHandleWithStandardInput];
NSData *inputData = [inputFile availableData];
NSString *word = [[NSString alloc]initWithData:inputData encoding:NSUTF8StringEncoding];
I want to convert a string into a double and after doing some math on it, convert it back to a string.
How do I do this in Objective-C?
Is there a way to round a double to the nearest integer too?
You can convert an NSString into a double with
double myDouble = [myString doubleValue];
Rounding to the nearest int can then be done as
int myInt = (int)(myDouble + (myDouble>0 ? 0.5 : -0.5))
I'm honestly not sure if there's a more streamlined way to convert back into a string than
NSString* myNewString = [NSString stringWithFormat:#"%d", myInt];
To really convert from a string to a number properly, you need to use an instance of NSNumberFormatter configured for the locale from which you're reading the string.
Different locales will format numbers differently. For example, in some parts of the world, COMMA is used as a decimal separator while in others it is PERIOD — and the thousands separator (when used) is reversed. Except when it's a space. Or not present at all.
It really depends on the provenance of the input. The safest thing to do is configure an NSNumberFormatter for the way your input is formatted and use -[NSFormatter numberFromString:] to get an NSNumber from it. If you want to handle conversion errors, you can use -[NSFormatter getObjectValue:forString:range:error:] instead.
Adding to olliej's answer, you can convert from an int back to a string with NSNumber's stringValue:
[[NSNumber numberWithInt:myInt] stringValue]
stringValue on an NSNumber invokes descriptionWithLocale:nil, giving you a localized string representation of value. I'm not sure if [NSString stringWithFormat:#"%d",myInt] will give you a properly localized reprsentation of myInt.
Here's a working sample of NSNumberFormatter reading localized number String (xCode 3.2.4, osX 10.6), to save others the hours I've just spent messing around. Beware: while it can handle trailing blanks such as "8,765.4 ", this cannot handle leading white space and this cannot handle stray text characters. (Bad input strings: " 8" and "8q" and "8 q".)
NSString *tempStr = #"8,765.4";
// localization allows other thousands separators, also.
NSNumberFormatter * myNumFormatter = [[NSNumberFormatter alloc] init];
[myNumFormatter setLocale:[NSLocale currentLocale]]; // happen by default?
[myNumFormatter setFormatterBehavior:NSNumberFormatterBehavior10_4];
// next line is very important!
[myNumFormatter setNumberStyle:NSNumberFormatterDecimalStyle]; // crucial
NSNumber *tempNum = [myNumFormatter numberFromString:tempStr];
NSLog(#"string '%#' gives NSNumber '%#' with intValue '%i'",
tempStr, tempNum, [tempNum intValue]);
[myNumFormatter release]; // good citizen
olliej's rounding method is wrong for negative numbers
2.4 rounded is 2 (olliej's method gets this right)
−2.4 rounded is −2 (olliej's method returns -1)
Here's an alternative
int myInt = (int)(myDouble + (myDouble>0 ? 0.5 : -0.5))
You could of course use a rounding function from math.h
// Converting String in to Double
double doubleValue = [yourString doubleValue];
// Converting Double in to String
NSString *yourString = [NSString stringWithFormat:#"%.20f", doubleValue];
// .20f takes the value up to 20 position after decimal
// Converting double to int
int intValue = (int) doubleValue;
or
int intValue = [yourString intValue];
For conversion from a number to a string, how about using the new literals syntax (XCode >= 4.4), its a little more compact.
int myInt = (int)round( [#"1.6" floatValue] );
NSString* myString = [#(myInt) description];
(Boxes it up as a NSNumber and converts to a string using the NSObjects' description method)
For rounding, you should probably use the C functions defined in math.h.
int roundedX = round(x);
Hold down Option and double click on round in Xcode and it will show you the man page with various functions for rounding different types.
This is the easiest way I know of:
float myFloat = 5.3;
NSInteger myInt = (NSInteger)myFloat;
from this example here, you can see the the conversions both ways:
NSString *str=#"5678901234567890";
long long verylong;
NSRange range;
range.length = 15;
range.location = 0;
[[NSScanner scannerWithString:[str substringWithRange:range]] scanLongLong:&verylong];
NSLog(#"long long value %lld",verylong);
convert text entered in textfield to integer
double mydouble=[_myTextfield.text doubleValue];
rounding to the nearest double
mydouble=(round(mydouble));
rounding to the nearest int(considering only positive values)
int myint=(int)(mydouble);
converting from double to string
myLabel.text=[NSString stringWithFormat:#"%f",mydouble];
or
NSString *mystring=[NSString stringWithFormat:#"%f",mydouble];
converting from int to string
myLabel.text=[NSString stringWithFormat:#"%d",myint];
or
NSString *mystring=[NSString stringWithFormat:#"%f",mydouble];
I ended up using this handy macro:
#define STRING(value) [#(value) stringValue]