Convert really large decimal string to hex? - objective-c

Question

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?

Answer 1

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.