CODEWITHSUNDEEP
iphoneobjective-c
Suppi
Suppi

Reputation: 630

How to convert Hex to Binary iphone

I need to convert a hex string to binary form in objective-c, Could someone please guide me? For example if i have a hex string 7fefff78, i want to convert it to 1111111111011111111111101111000?

BR, Suppi

Upvotes: 7

Views: 6278

Answers (4)

Hardik Darji
Hardik Darji

Reputation: 3684

I agree with kerrek SB's answer and tried this. Its work for me.

+(NSString *)convertBinaryToHex:(NSString *) strBinary
{
    NSString *strResult = @"";
    NSDictionary *dictBinToHax = [[NSDictionary alloc] initWithObjectsAndKeys:
                                  @"0",@"0000",
                                  @"1",@"0001",
                                  @"2",@"0010",
                                  @"3",@"0011",

                                  @"4",@"0100",
                                  @"5",@"0101",
                                  @"6",@"0110",
                                  @"7",@"0111",

                                  @"8",@"1000",
                                  @"9",@"1001",
                                  @"A",@"1010",
                                  @"B",@"1011",

                                  @"C",@"1100",
                                  @"D",@"1101",
                                  @"E",@"1110",
                                  @"F",@"1111", nil];

    for (int i = 0;i < [strBinary length]; i+=4)
    {
        NSString *strBinaryKey = [strBinary substringWithRange: NSMakeRange(i, 4)];
        strResult = [NSString stringWithFormat:@"%@%@",strResult,[dictBinToHax valueForKey:strBinaryKey]];
    }
    return  strResult;
}

Upvotes: 0

Karu Ma
Karu Ma

Reputation: 323

In case you need leading zeros, for example 18 returns 00011000 instead of 11000

-(NSString *)toBinary:(NSUInteger)input strLength:(int)length{
        if (input == 1 || input == 0){

         NSString *str=[NSString stringWithFormat:@"%u", input];
            return str;
        }
        else {
            NSString *str=[NSString stringWithFormat:@"%@%u", [self toBinary:input / 2 strLength:0], input % 2];
            if(length>0){
                int reqInt = length * 4;
                for(int i= [str length];i < reqInt;i++){
                    str=[NSString stringWithFormat:@"%@%@",@"0",str];
                }
            }
            return str;
        }  
}
 NSString *hex = @"58";
 NSUInteger hexAsInt;
 [[NSScanner scannerWithString:hex] scanHexInt:&hexAsInt];
 NSString *binary = [NSString stringWithFormat:@"%@", [self toBinary:hexAsInt strLength:[hex length]]];
NSLog(@"binario %@",binary);

Upvotes: 1

Mundi
Mundi

Reputation: 80265

Nice recursive solution...

NSString *hex = @"49cf3e";
NSUInteger hexAsInt;
[[NSScanner scannerWithString:hex] scanHexInt:&hexAsInt];
NSString *binary = [NSString stringWithFormat:@"%@", [self toBinary:hexAsInt]];

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

Upvotes: 8

Kerrek SB
Kerrek SB

Reputation: 477040

Simply convert each digit one by one: 0 -> 0000, 7 -> 0111, F -> 1111, etc. A little lookup table could make this very concise.

The beauty of number bases that are powers of another base :-)

Upvotes: 2

Related Questions