如何将十六进制字符串转换为十六进制数组？

Question

例如：
我有一个字符串：hexInString = @"0123456789abcdef"，我想将它转换为十六进制数组，如：

{01, 23, 45, 67, 89, ab, cd, ef}

或：

{0123, 4567, 89ab, cdef}

我尝试这样做，但因十进制数组而失败：

+ (NSMutableArray *)generateHexArrayFrom:(NSString *)hexInString
{
  NSMutableArray * hexArray = [[NSMutableArray alloc] init];
  int hexStringSize = sizeof(uint_fast32_t) / 2;

  for (int x = 0; x <= [hexInString length] - hexStringSize; x += hexStringSize) {
    NSScanner * scanner = [[NSScanner alloc] initWithString:[hexInString substringWithRange:NSMakeRange(x, hexStringSize)]];
    uint_fast32_t hex;
    [scanner scanHexInt:&hex];
    [hexArray addObject:[NSNumber numberWithInt:hex]];
  }

  return [hexArray autorelease];
}

如何向NSMutableArray添加十六进制值？或者是否有存储数据的类型？