Converting NSString to char byte array - objective-c

Question

In my program, I receive a NSString like this one : AA010158AA7D385002. And I need to pass it to a method which accept a char byte array, as below :
char[9] = {0xAA, 0x01, 0x01, 0x58, 0xAA, 0x7D, 0x38, 0x50, 0x02};
How to convert my NSString to char byte array like this one?
Thanks!

Answer 1

NSString *strCharData = #"AA010158AA7D385002";
const char *characterRes = [strCharData cStringUsingEncoding:NSUTF8StringEncoding];
or
NSString *strCharData = #"AA010158AA7D385002";
const char *characterRes = [strCharData UTF8String];

Answer 2

Use this answer if i am correct,i did little coding but might be there are possibilities of simpler solutions also like #user3182143
NSString * inputStr = #"AA010158AA7D385002";
NSMutableArray *charByteArray = [[NSMutableArray alloc]initWithCapacity:1];
int i = 0;
for (i = 0; i+2 <= inputStr.length; i+=2) {
NSRange range = NSMakeRange(i, 2);
NSString* charStr = [inputStr substringWithRange:range];
[charByteArray addObject:[NSString stringWithFormat:#"0x%#",charStr]];
}
Output :
char[9] = (
0xAA,
0x01,
0x01,
0x58,
0xAA,
0x7D,
0x38,
0x50,
0x02
)

Answer 3

Since your text is hex and you want actual bytes out (which each correspond to two hex characters), you'll have to manually convert each character into the corresponding number, and then add them together into the correct numerical value.
To do this, I'm taking advantage of the fact that, in ASCII characters, a...z are in a row, as are 0...9. I'm also taking advantage of the fact that hexadecimal is valid ASCII, and that Unicode characters from 0...127 are identical to their corresponding ASCII characters.
Below is a program that does this and prints out the original string's characters as well as the calculated bytes (as hex again):
#import <Foundation/Foundation.h>
int main(int argc, char *argv[])
{
#autoreleasepool
{
NSString *hexStr = #"1234567890abcdef12";
unsigned char theBytes[9] = {};
for( NSUInteger x = 0; x < sizeof(theBytes); x++ )
{
unsigned char digitOne = [hexStr characterAtIndex: x * 2];
if( digitOne >= 'a' )
digitOne -= 'a' -10;
else
digitOne -= '0';
unsigned char digitTwo = [hexStr characterAtIndex: (x * 2) +1];
if( digitTwo >= 'a' )
digitTwo -= 'a' -10;
else
digitTwo -= '0';
printf("%01x%01x",digitOne,digitTwo);
theBytes[x] = (digitOne << 4) | digitTwo;
}
printf("\n");
for( int x = 0; x < sizeof(theBytes); x++ )
printf("%02x",theBytes[x]);
printf("\n");
}
}
Note: This code naïvely assumes that you are providing a correct string. I.e. your input has to consist of lowercase characters and numbers only, and exactly 18 characters. Anything else and you get a wrong result or an exception.

Answer 4

I finally managed to find the answer to my own question. I am posting it here in case it helps someone else.
So I use a method to convert an NSString hex to bytes :
#implementation NSString (HexToBytes)
- (NSData *)hexToBytes
{
NSMutableData *data = [NSMutableData data];
int idx;
for (idx = 0; idx + 2 <= self.length; idx += 2) {
NSRange range = NSMakeRange(idx, 2);
NSString *hexStr = [self substringWithRange:range];
NSScanner *scanner = [NSScanner scannerWithString:hexStr];
unsigned int intValue;
[scanner scanHexInt:&intValue];
[data appendBytes:&intValue length:1];
}
return data;
}
#end
And then, I simply use it like that :
NSString *str = #"AA010158AA7D385002";
NSData *databytes = [str hexToBytes];
char *bytePtr = (char *)[databytes bytes];
And I finally get my char array. Hope it helps someone else.