Convert Hexadecimal to Binary (large values)

Question

-(NSString *)toBinary:(NSUInteger)input
{
    if (input == 1 || input == 0)
        return [NSString stringWithFormat:@"%u", input];
    return [NSString stringWithFormat:@"%@%u", [self toBinary:input / 2], input % 2];
}

NSString *hex = txtHexInput.text;
NSUInteger hexAsInt;
[[NSScanner scannerWithString:hex] scanHexInt:&hexAsInt];
NSString *binary = [NSString stringWithFormat:@"%@", [self toBinary:hexAsInt]];
txtBinaryInput.text = binary;

The above code works great... that is until you need to exceed 32 bits. Any pointers to converting hex to binary for larger than 32 bit values? Thank you.

Use `uint64_t` to get 64 bits. Beyond that you need to describe how you get larger integers. — rmaddy, Nov 29 '13 at 23:48
I'm looking into this now... is there an efficient way to incorporate that into my current code? — landfill, Nov 30 '13 at 00:15

rmaddy · Answer 1 · 2013-11-30T00:38:50.137

2

You can get 64 bits using uint64_t or unsigned long long.

-(NSString *)toBinary:(unsigned long long)input
{
    if (input == 1 || input == 0)
        return [NSString stringWithFormat:@"%llu", input];
    return [NSString stringWithFormat:@"%@%llu", [self toBinary:input / 2], input % 2];
}

NSString *hex = txtHexInput.text;
unsigned long long hexAsULL;
[[NSScanner scannerWithString:hex] scanHexLongLong:&hexAsULL];
NSString *binary = [NSString stringWithFormat:@"%@", [self toBinary:hexAsULL]];
txtBinaryInput.text = binary;

This will give you numbers from 0 to 18,446,744,073,709,551,615 (decimal)

edited Nov 30 '13 at 00:38

answered Nov 30 '13 at 00:28

rmaddy

314,917
42
532
579

Oh thank you! I was working along those lines but did not finish it off correctly. I changed the %ull to %llu. – landfill Nov 30 '13 at 00:36

Convert Hexadecimal to Binary (large values)

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