How to convert the Char * to hexadecimal value - iphone

Question

I have a char array that i need to convert to hexadecimal value
char* arrrr =[self mountLVparams:NULL :c :code_ward_arr];
int size =strlen(arrrr);
I am trying with this but this not happening
NSData* data = [NSData dataWithBytes:arrrr length:sizeof(sizeof(unsigned char)*size)];

Answer 1

Try this:
NSString *inp = [NSString stringWithString:#"MmMmMm"];
char star = [inp characterAtIndex:3];
NSString *string = [NSString stringWithFormat:#"char:%c andHex:%x",[inp characterAtIndex:3],star];
NSLog(#"%#",string);
Look at in output is "%#" as a regular string. In string you will receive hex value of char.
char:m andHex:6d
With array you will have to do it in a loop

Answer 2

You don't "convert" a value to hexadecimal. Hexadecimal is just a different way of counting or displaying the same numbers.
if (0xFF==255)
{
NSLog(#"255 equals 0xFF");
}
Now see the following code below which will convert your byte array into and int, then display then format the int as a hex string.
-(void) convertData: (unsigned char*) data
{
int numBytes = sizeof(int);
//convert array of bytes to NSData*
NSData* binaryData = [[NSData alloc] initWithBytes: (void*) data length: numBytes];
//read value of number stored in data, presuming it is an integer
int number = *((int*) data);
//format number as a hex string
NSString* hexString = [NSString stringWithFormat:#"%X",number];
}