I trying to do conversions between Binary, Octal, Decimal and Hexadecimal in Objective-C.
I had problems converting Octal to Decimal.
I have tried the following:
NSString *decString = [NSString stringWithFormat:#"%d", 077];
It works fine, returning 63 as expected, but my Octal value is a NSString. How can I tell the computer that it is a Octal;
I know there is a method called "scanHexInt:" which I used to convert Hexadecimal to decimal, but it seems there is no scanOctInt...
Any help would be appreciated!
The cleanest solution is probably:
long result = strtol(input.UTF8String, NULL, 8);
or
long long result = strtoll(input.UTF8String, NULL, 8);
Define a category on NSString (put this on top of any of your source code modules or into a new .m/.h file pair, #interface goes into .h, #implementation into .m):
#interface NSString (NSStringWithOctal)
-(int)octalIntValue;
#end
#implementation NSString (NSStringWithOctal)
-(int)octalIntValue
{
int iResult = 0, iBase = 1;
char c;
for(int i=(int)[self length]-1; i>=0; i--)
{
c = [self characterAtIndex:i];
if((c<'0')||(c>'7')) return 0;
iResult += (c - '0') * iBase;
iBase *= 8;
}
return iResult;
}
#end
Use it like that:
NSString *s = #"77";
int i = [s octalIntValue];
NSLog(#"%d", i);
The method returns an integer representing the octal value in the string. It returns 0, if the string is not an octal number. Leading zeroes are allowed, but not necessary.
Alternatively, if you want to drop down to C, you can use sscanf
int oct;
sscanf( [yourString UTF8String], "%o", &oct );
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
I've got a really large decimal number in an NSString, which is too large to fit into any variable including NSDecimal. I was doing the math manually, but if I can't fit the number into a variable then I can't be dividing it. So what would be a good way to convert the string?
Example Input: 423723487924398723478243789243879243978234
Output: 4DD361F5A772159224CE9EB0C215D2915FA
I was looking at the first answer here, but it's in C# and I don't know it's objective C equivalent.
Does anyone have any ideas that don't involve using an external library?
If this is all you need, it's not too hard to implement, especially if you're willing to use Objective-C++. By using Objective-C++, you can use a vector to manage memory, which simplifies the code.
Here's the interface we'll implement:
// NSString+BigDecimalToHex.h
#interface NSString (BigDecimalToHex)
- (NSString *)hexStringFromDecimalString;
#end
To implement it, we'll represent an arbitrary-precision non-negative integer as a vector of base-65536 digits:
// NSString+BigDecimalToHex.mm
#import "NSString+BigDecimalToHex.h"
#import <vector>
// index 0 is the least significant digit
typedef std::vector<uint16_t> BigInt;
The "hard" part is to multiply a BigInt by 10 and add a single decimal digit to it. We can very easily implement this as long multiplication with a preloaded carry:
static void insertDecimalDigit(BigInt &b, uint16_t decimalDigit) {
uint32_t carry = decimalDigit;
for (size_t i = 0; i < b.size(); ++i) {
uint32_t product = b[i] * (uint32_t)10 + carry;
b[i] = (uint16_t)product;
carry = product >> 16;
}
if (carry > 0) {
b.push_back(carry);
}
}
With that helper method, we're ready to implement the interface. First, we need to convert the decimal digit string to a BigInt by calling the helper method once for each decimal digit:
- (NSString *)hexStringFromDecimalString {
NSUInteger length = self.length;
unichar decimalCharacters[length];
[self getCharacters:decimalCharacters range:NSMakeRange(0, length)];
BigInt b;
for (NSUInteger i = 0; i < length; ++i) {
insertDecimalDigit(b, decimalCharacters[i] - '0');
}
If the input string is empty, or all zeros, then b is empty. We need to check for that:
if (b.size() == 0) {
return #"0";
}
Now we need to convert b to a hex digit string. The most significant digit of b is at the highest index. To avoid leading zeros, we'll handle that digit specially:
NSMutableString *hexString = [NSMutableString stringWithFormat:#"%X", b.back()];
Then we convert each remaining base-65536 digit to four hex digits, in order from most significant to least significant:
for (ssize_t i = b.size() - 2; i >= 0; --i) {
[hexString appendFormat:#"%04X", b[i]];
}
And then we're done:
return hexString;
}
You can find my full test program (to run as a Mac command-line program) in this gist.
The following is tried to print out N number of spaces (or 12 in the example):
NSLog(#"hello%#world", [NSString stringWithCharacters:" " length:12]);
const unichar arrayChars[] = {' '};
NSLog(#"hello%#world", [NSString stringWithCharacters:arrayChars length:12]);
const unichar oneChar = ' ';
NSLog(#"hello%#world", [NSString stringWithCharacters:&oneChar length:12]);
But they all print out weird things such as hello ÔÅÓñüÔÅ®Óñü®ÓüÅ®ÓñüÔ®ÓüÔÅ®world... I thought a "char array" is the same as a "string" and the same as a "pointer to a character"? The API spec says it is to be a "C array of Unicode characters" (by Unicode, is it UTF8? if it is, then it should be compatible with ASCII)... How to make it work and why those 3 ways won't work?
You can use %*s to specify the width.
NSLog(#"Hello%*sWorld", 12, "");
Reference:
A field width, or precision, or both, may be indicated by an asterisk
( '*' ). In this case an argument of type int supplies the field width
or precision. Applications shall ensure that arguments specifying
field width, or precision, or both appear in that order before the
argument, if any, to be converted.
This will get you what you want:
NSLog(#"hello%#world", [#"" stringByPaddingToLength:12 withString:#" " startingAtIndex:0]);
I think the issue you have is you are misinterpreting what +(NSString *)stringWithCharacters:length: is supposed to do. It's not supposed to repeat the characters, but instead copy them from the array into a string.
So in your case you only have a single ' ' in the array, meaning the other 11 characters will be taken from whatever follows arrayChars in memory.
If you want to print out a pattern of n spaces, the easiest way to do that would be to use -(NSString *)stringByPaddingToLength:withString:startingAtIndex:, i.e creating something like this.
NSString *formatString = #"Hello%#World";
NSString *paddingString = [[NSString string] stringByPaddingToLength: n withString: #" " startingAtIndex: 0];
NSLog(formatString, paddingString);
This is probably the fastest method:
NSString *spacesWithLength(int nSpaces)
{
char UTF8Arr[nSpaces + 1];
memset(UTF8Arr, ' ', nSpaces * sizeof(*UTF8Arr));
UTF8Arr[nSpaces] = '\0';
return [NSString stringWithUTF8String:UTF8Arr];
}
The reason your current code isn't working is because +stringWithCharacters: expects an array with a length of characters of 12, while your array is only 1 character in length {' '}. So, to fix, you must create a buffer for your array (in this case, we use a char array, not a unichar, because we can easily memset a char array, but not a unichar array).
The method I provided above is probably the fastest that is possible with a dynamic length. If you are willing to use GCC extensions, and you have a fixed size array of spaces you need, you can do this:
NSString *spacesWithLength7()
{
unichar characters[] = { [0 ... 7] = ' ' };
return [NSString stringWithCharacters:characters length:7];
}
Unfortunately, that extension doesn't work with variables, so it must be a constant.
Through the magic of GCC extensions and preprocessor macros, I give you.... THE REPEATENATOR! Simply pass in a string (or a char), and it will do the rest! Buy now, costs you only $19.95, operators are standing by! (Based on the idea suggested by #JeremyL)
// step 1: determine if char is a char or string, or NSString.
// step 2: repeat that char or string
// step 3: return that as a NSString
#define repeat(inp, cnt) __rep_func__(#encode(typeof(inp)), inp, cnt)
// arg list: (int siz, int / char *input, int n)
static inline NSString *__rep_func__(char *typ, ...)
{
const char *str = NULL;
int n;
{
va_list args;
va_start(args, typ);
if (typ[0] == 'i')
str = (const char []) { va_arg(args, int), '\0' };
else if (typ[0] == '#')
str = [va_arg(args, id) UTF8String];
else
str = va_arg(args, const char *);
n = va_arg(args, int);
va_end(args);
}
int len = strlen(str);
char outbuf[(len * n) + 1];
// now copy the content
for (int i = 0; i < n; i++) {
for (int j = 0; j < len; j++) {
outbuf[(i * len) + j] = str[j];
}
}
outbuf[(len * n)] = '\0';
return [NSString stringWithUTF8String:outbuf];
}
The stringWithCharaters:length: method makes an NSString (or an instance of a subclass of NSString) using the first length characters in the C array. It does not iterate over the given array of characters until it reaches the length.
The output you are seeing is the area of memory 12 Unicode characters long starting at the location of your passed 1 Unicode character array.
This should work.
NSLog(#"hello%#world", [NSString stringWithCharacters:" " length:12]);
How can I enumerate NSString by pulling each unichar out of it? I can use characterAtIndex but that is slower than doing it by an incrementing unichar*. I didn't see anything in Apple's documentation that didn't require copying the string into a second buffer.
Something like this would be ideal:
for (unichar c in string) { ... }
or
unichar* ptr = (unichar*)string;
You can speed up -characterAtIndex: by converting it to it's IMP form first:
NSString *str = #"This is a test";
NSUInteger len = [str length]; // only calling [str length] once speeds up the process as well
SEL sel = #selector(characterAtIndex:);
// using typeof to save my fingers from typing more
unichar (*charAtIdx)(id, SEL, NSUInteger) = (typeof(charAtIdx)) [str methodForSelector:sel];
for (int i = 0; i < len; i++) {
unichar c = charAtIdx(str, sel, i);
// do something with C
NSLog(#"%C", c);
}
EDIT: It appears that the CFString Reference contains the following method:
const UniChar *CFStringGetCharactersPtr(CFStringRef theString);
This means you can do the following:
const unichar *chars = CFStringGetCharactersPtr((__bridge CFStringRef) theString);
while (*chars)
{
// do something with *chars
chars++;
}
If you don't want to allocate memory for coping the buffer, this is the way to go.
Your only option is to copy the characters into a new buffer. This is because the NSString class does not guarantee that there is an internal buffer you can use. The best way to do this is to use the getCharacters:range: method.
NSUInteger i, length = [string length];
unichar *buffer = malloc(sizeof(unichar) * length);
NSRange range = {0,length};
[string getCharacters:buffer range:range];
for(i = 0; i < length; ++i) {
unichar c = buffer[i];
}
If you are using potentially very long strings, it would be better to allocate a fixed size buffer and enumerate the string in chunks (this is actually how fast enumeration works).
I created a block-style enumeration method that uses getCharacters:range: with a fixed-size buffer, as per ughoavgfhw's suggestion in his answer. It avoids the situation where CFStringGetCharactersPtr returns null and it doesn't have to malloc a large buffer. You can drop it into an NSString category, or modify it to take a string as a parameter if you like.
-(void)enumerateCharactersWithBlock:(void (^)(unichar, NSUInteger, BOOL *))block
{
const NSInteger bufferSize = 16;
const NSInteger length = [self length];
unichar buffer[bufferSize];
NSInteger bufferLoops = (length - 1) / bufferSize + 1;
BOOL stop = NO;
for (int i = 0; i < bufferLoops; i++) {
NSInteger bufferOffset = i * bufferSize;
NSInteger charsInBuffer = MIN(length - bufferOffset, bufferSize);
[self getCharacters:buffer range:NSMakeRange(bufferOffset, charsInBuffer)];
for (int j = 0; j < charsInBuffer; j++) {
block(buffer[j], j + bufferOffset, &stop);
if (stop) {
return;
}
}
}
}
The fastest reliable way to enumerate characters in an NSString I know of is to use this relatively little-known Core Foundation gem hidden in plain sight (CFString.h).
NSString *string = <#initialize your string#>
NSUInteger stringLength = string.length;
CFStringInlineBuffer buf;
CFStringInitInlineBuffer((__bridge CFStringRef) string, &buf, (CFRange) { 0, stringLength });
for (NSUInteger charIndex = 0; charIndex < stringLength; charIndex++) {
unichar c = CFStringGetCharacterFromInlineBuffer(&buf, charIndex);
}
If you look at the source code of these inline functions, CFStringInitInlineBuffer() and CFStringGetCharacterFromInlineBuffer(), you'll see that they handle all the nasty details like CFStringGetCharactersPtr() returning NULL, CFStringGetCStringPtr() returning NULL, defaulting to slower CFStringGetCharacters() and caching the characters in a C array for fastest access possible. This API really deserves more publicity.
The caveat is that if you initialize the CFStringInlineBuffer at a non-zero offset, you should pass a relative character index to CFStringInlineBuffer(), as stated in the header comments:
The next two functions allow fast access to the contents of a string, assuming you are doing sequential or localized accesses. To use, call CFStringInitInlineBuffer() with a CFStringInlineBuffer (on the stack, say), and a range in the string to look at. Then call CFStringGetCharacterFromInlineBuffer() as many times as you want, with a index into that range (relative to the start of that range). These are INLINE functions and will end up calling CFString only once in a while, to fill a buffer. CFStringGetCharacterFromInlineBuffer() returns 0 if a location outside the original range is specified.
I don't think you can do this. NSString is an abstract interface to a multitude of classes that make no guarantees about the internal storage of the character data, so it's entirely possible there is no character array to get a pointer to.
If neither of the options mentioned in your question are suitable for your app, I'd recommend either creating your own string class for this purpose, or using raw malloc'ed unichar arrays instead of string objects.
This will work:
char *s = [string UTF8String];
for (char *t = s; *t; t++)
/* use as */ *t;
[Edit] And if you really need unicode characters then you have no option but to use length and characterAtIndex. From the documentation:
The NSString class has two primitive methods—length and characterAtIndex:—that provide the basis for all other methods in its interface. The length method returns the total number of Unicode characters in the string. characterAtIndex: gives access to each character in the string by index, with index values starting at 0.
So your code would be:
for (int index = 0; index < string.length; index++)
{
unichar c = [string characterAtIndex: index];
/* ... */
}
[edit 2]
Also, don't forget that NSString is 'toll-free bridged' to CFString and thus all the non-Objective-C, straight C-code interface functions are usable. The relevant one would be CFStringGetCharacterAtIndex
code below.
i'm tryind to obtain string answers like "a1", "c4"
this is what i'm having instead of "a1": "adresse finale: \340}00\214"
with this prinf:
printf("\nadresse finale: %s",[self convertCGPointToSquareAdress:self.frame.origin]);
the method is:
-(NSString *) convertCGPointToSquareAdress:(CGPoint ) point{
int x= point.x /PIECE_WIDTH;
int y=point.y/PIECE_WIDTH;
char lettreChiffre[2];
//char chiffre;
NSString *squareAdress;
//ascii a=97 , b=98... h=105
for (int i=97; i<105; i++) {
for (int j=8; j>0; j--) {
if(i-97==x && j-1==y ){
NSLog(#"enterrrrrrrrrred if convertCGPointToSquareAdress");
lettreChiffre[0]=i;
lettreChiffre[1]=(char) j;
printf(" lettreChiffre: %s ", lettreChiffre);
NSString *squareAdress=[NSString stringWithFormat:#"%s", lettreChiffre];
break;
}
}
}
return squareAdress;
}
can you please help me?
thanks in advance.
There are three problems I can see with your code:
1.
When you do
lettreChiffre[1]=(char) j;
remember j is a number between 1 and 8, so you're getting the ASCII character whose value is j, not the character 1...8. You should use
lettreChiffre[1]= '0' + j;
2.
lettreChiffre is a char array of length 2, which means there's no room for the terminal null character. This may work, but may give you gibberish. You should instead declare
char lettreChiffre[3];
lettreChiffre[2] = '\0';
3.
You're trying to use printf to print an NSString, which it can't do. Either use
NSLog(#"adresse finale: %#", mynsstring)
or convert the NSString back to a C-string:
printf("adresse finale: %s", [mynsstring UTF8String]);
Also, as noted by #dreamlax, you don't really need the loop. I assumed you were doing something else and ran into this trouble, so we're not really seeing the full code. But, if this is really the entirety of your code, then you can simply remove the loop as #dreamlax suggested.
What is the purpose of the loop? You have a loop that essentially brute forces a matrix to calculate the “square address”. Your method will also return an uninitialized pointer if x is greater than 8.
Your entire method could be made much simpler.
- (NSString *) convertCGPointToSquareAdress:(CGRect) point
{
unsigned int x = point.x / PIECE_WIDTH;
unsigned int y = point.y / PIECE_WIDTH;
// Do some range checking to ensure x and y are valid.
char lettreChiffre[3];
lettreChiffre[0] = 'a' + x;
lettreChiffre[1] = '1' + y;
lettreChiffre[2] = '\0';
return [NSString stringWithCString:letterChiffre encoding:NSASCIIStringEncoding];
}