Base 62 conversion in Objective-C - ios

Question

I spent much too much time trying to find an implementation for base 62 conversion for Objective-C. I am sure this is a terrible example and there must be an elegant, super-efficient way to do this, but this works, please edit or answer to improve it! But I wanted to help people searching for this to have something that will work. There doesn't appear to be anything specific to be found for an Objective-C implementation.
#implementation Base62Converter
+(int)decode:(NSString*)string
{
int num = 0;
NSString * alphabet = #"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
for (int i = 0, len = [string length]; i < len; i++)
{
NSRange range = [alphabet rangeOfString:[string substringWithRange:NSMakeRange(i,1)]];
num = num * 62 + range.location;
}
return num;
}
+(NSString*)encode:(int)num
{
NSString * alphabet = #"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
NSMutableString * precursor = [NSMutableString stringWithCapacity:3];
while (num > 0)
{
[precursor appendString:[alphabet substringWithRange:NSMakeRange( num % 62, 1 )]];
num /= 62;
}
// http://stackoverflow.com/questions/6720191/reverse-nsstring-text
NSMutableString *reversedString = [NSMutableString stringWithCapacity:[precursor length]];
[precursor enumerateSubstringsInRange:NSMakeRange(0,[precursor length])
options:(NSStringEnumerationReverse |NSStringEnumerationByComposedCharacterSequences)
usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
[reversedString appendString:substring];
}];
return reversedString;
}
#end

Answer 1

Your code is fine. If anything, make it more generic. Here is a recursive version for any base (same code):
#import <Foundation/Foundation.h>
#interface BaseConversion : NSObject
+(NSString*) formatNumber:(NSUInteger)n toBase:(NSUInteger)base;
+(NSString*) formatNumber:(NSUInteger)n usingAlphabet:(NSString*)alphabet;
#end
#implementation BaseConversion
// Uses the alphabet length as base.
+(NSString*) formatNumber:(NSUInteger)n usingAlphabet:(NSString*)alphabet
{
NSUInteger base = [alphabet length];
if (n<base){
// direct conversion
NSRange range = NSMakeRange(n, 1);
return [alphabet substringWithRange:range];
} else {
return [NSString stringWithFormat:#"%#%#",
// Get the number minus the last digit and do a recursive call.
// Note that division between integer drops the decimals, eg: 769/10 = 76
[self formatNumber:n/base usingAlphabet:alphabet],
// Get the last digit and perform direct conversion with the result.
[alphabet substringWithRange:NSMakeRange(n%base, 1)]];
}
}
+(NSString*) formatNumber:(NSUInteger)n toBase:(NSUInteger)base
{
NSString *alphabet = #"0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"; // 62 digits
NSAssert([alphabet length]>=base,#"Not enough characters. Use base %ld or lower.",(unsigned long)[alphabet length]);
return [self formatNumber:n usingAlphabet:[alphabet substringWithRange:NSMakeRange (0, base)]];
}
#end
int main(int argc, char *argv[]) {
#autoreleasepool {
NSLog(#"%#",[BaseConversion formatNumber:3735928559 toBase:16]); // deadbeef
return EXIT_SUCCESS;
}
}
A Swift 3 version: https://gist.github.com/j4n0/056475333d0ddfe963ac5dc44fa53bf2

Answer 2

You could improve your encode method in such a way that reversing the final string is not necessary:
+ (NSString *)encode:(NSUInteger)num
{
NSString *alphabet = #"0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
NSUInteger base = [alphabet length];
NSMutableString *result = [NSMutableString string];
while (num > 0) {
NSString *digit = [alphabet substringWithRange:NSMakeRange(num % base, 1)];
[result insertString:digit atIndex:0];
num /= base;
}
return result;
}
Of course, this could also be generalized for arbitrary bases or alphabets, as suggested by #Jano in his answer.
Note that this method (as well as your original encode method) returns an empty string for num = 0, so you might want to consider this case separately (or just replace while (num > 0) { ... } by do { ... } while (num > 0).
For more efficiency, one could avoid all intermediate NSString objects altogether, and work with plain C strings:
+ (NSString *)encode:(NSUInteger)num
{
static const char *alphabet = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
NSUInteger base = 62;
char result[20]; // sufficient room to encode 2^64 in Base-62
char *p = result + sizeof(result);
*--p = 0; // NULL termination
while (num > 0) {
*--p = alphabet[num % base];
num /= base;
}
return [NSString stringWithUTF8String:p];
}