Are there any methods in objective C for converting byte to int, float and NSString?
The first are C-types. No conversion is needed, just assign them:
byte b = ...;
int x = b;
float f = b;
Converting to NSString could be done using stringWithFormat:, a NSNumberFormatter and many more methods. This is the easiest:
NSString *myString = [NSString stringWithFormat: #"%d", b];
If you want it printed in hex, use #"%x" (for lowercase letters) or #"%X" (for capital letters) instead.
You could get an NSString from NSData with initWithData:encoding: and then transform it into int or float by calling intValue and floatValue
Related
I am trying to convert a nsstring with hex values into a float value.
NSString *hexString = #"3f9d70a4";
The float value should be = 1.230.
Some ways I have tried to solve this are:
1.NSScanner
-(unsigned int)strfloatvalue:(NSString *)str
{
float outVal;
NSString *newStr = [NSString stringWithFormat:#"0x%#",str];
NSScanner* scanner = [NSScanner scannerWithString:newStr];
NSLog(#"string %#",newStr);
bool test = [scanner scanHexFloat:&outVal];
NSLog(#"scanner result %d = %a (or %f)",test,outVal,outVal);
return outVal;
}
results:
string 0x3f9d70a4
scanner result 1 = 0x1.fceb86p+29 (or 1067282624.000000)
2.casting pointers
NSNumber * xPtr = [NSNumber numberWithFloat:[(NSNumber *)#"3f9d70a4" floatValue]];
result:3.000000
What you have is not a "hexadecimal float", as is produced by the %a string format and scanned by scanHexFloat: but the hexadecimal representation of a 32-bit floating-point value - i.e. the actual bits.
To convert this back to a float in C requires messing with the type system - to give you access to the bytes that make up a floating-point value. You can do this with a union:
typedef union { float f; uint32_t i; } FloatInt;
This type is similar to a struct but the fields are overlaid on top of each other. You should understand that doing this kind of manipulation requires you understand the storage formats, are aware of endian order, etc. Do not do this lightly.
Now you have the above type you can scan a hexadecimal integer and interpret the resultant bytes as a floating-point number:
FloatInt fl;
NSScanner *scanner = [NSScanner scannerWithString:#"3f9d70a4"];
if([scanner scanHexInt:&fl.i]) // scan into the i field
{
NSLog(#"%x -> %f", fl.i, fl.f); // display the f field, interpreting the bytes of i as a float
}
else
{
// parse error
}
This works, but again consider carefully what you are doing.
HTH
I think a better solutions is a workaround like this :
-(float) getFloat:(NSInteger*)pIndex
{
NSInteger index = *pIndex;
NSData* data = [self subDataFromIndex:&index withLength:4];
*pIndex = index;
uint32_t hostData = CFSwapInt32BigToHost(*(const uint32_t *)[data bytes]);
return *(float *)(&hostData);;
}
Where your parameter is an NSData which rapresents the number in HEX format, and the input parameter is a pointer to the element of NSData.
So basically you are trying to make an NSString to C's float, there's an old fashion way to do that!
NSString* hexString = #"3f9d70a4";
const char* cHexString = [hexString UTF8String];
long l = strtol(cHexString, NULL, 16);
float f = *((float *) &l);
// f = 1.23
for more detail please see this answer
just like my question, How can i convert decimal into octal in objective c?
can somebody help me? it's make me dizy
You have to initialize a new NSString object using the right format specifier.
Like :
int i = 9;
NSString *octalString = [NSString stringWithFormat:#"%o", i]; // %O works too.
This will create an autoreleased NSString object containing octal string representation of i. In case you start from a NSString
containing a decimal number representation, you should first retrieve the number using :
int i = [myDecimalNumberAsAString intValue];
Here is a link to the format specifiers reference.
Hope this helps.
Since Objective C is a superset of C you can just use C functions such as sprintf, e.g.
char s[32];
int n = 42;
sprintf(s, "%o", n);
How do I convert a char to an NSString in Objective-C?
Not a null-terminated C string, just a simple char c = 'a'.
You can use stringWithFormat:, passing in a format of %c to represent a character, like this:
char c = 'a';
NSString *s = [NSString stringWithFormat:#"%c", c];
You can make a C-string out of one character like this:
char cs[2] = {c, 0}; //c is the character to convert
NSString *s = [[NSString alloc] initWithCString:cs encoding: SomeEncoding];
Alternatively, if the character is known to be an ASCII character (i. e. Latin letter, number, or a punctuation sign), here's another way:
unichar uc = (unichar)c; //Just extend to 16 bits
NSString *s = [NSString stringWithCharacters:&uc length:1];
The latter snippet with surely fail (not crash, but produce a wrong string) with national characters. For those, simple extension to 16 bits is not a correct conversion to Unicode. That's why the encoding parameter is needed.
Also note that the two snippets above produce a string with diferent deallocation requirements. The latter makes an autoreleased string, the former makes a string that needs a [release] call.
Are there any Cocoa classes that will help me convert a hex value in a NSString like 0x12FA to a long or NSNumber? It doesn't look like any of the classes like NSNumberFormatter support hex numbers.
Thanks,
Hua-Ying
Here's a short example of how you would do it using NSScanner:
NSString* pString = #"0xDEADBABE";
NSScanner* pScanner = [NSScanner scannerWithString: pString];
unsigned int iValue;
[pScanner scanHexInt: &iValue];
See NSScanner's scanHex...: methods. That'll get you the primitive that you can wrap in an NSNumber.
here is the other way conversion, a long long int to hex string.
first the hex to long long.
NSString* pString = #"ffffb382ddfe";
NSScanner* pScanner = [NSScanner scannerWithString: pString];
unsigned long long iValue2;
[pScanner scanHexLongLong: &iValue2];
NSLog(#"iValue2 = %lld", iValue2);
and the other way, longlong to hex string...
NSNumber *number;
NSString *hexString;
number = [NSNumber numberWithLongLong:iValue2];
hexString = [NSString stringWithFormat:#"%qx", [number longLongValue]];
NSLog(#"hexString = %#", hexString);
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]