Creating Random Alpha numeric on IOS - ios

Question

i'm java programmer that 'must' move on to obj-C for a while,
i got some confuse when generating random alphanumeric code... here my javacode:
PS: i want to generate code like this :Gh12PU67, AC88pP13, Bk81gH89
private String generateCode(){
String code = "";
Random r = new Random();
char[] c = new char[]{'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
for(int i = 0; i<4; i++){
int uplow = r.nextInt(2);
String temp = ""+ c[r.nextInt(c.length)];
if(uplow==1)
code = code + temp.toUpperCase();
else
code = code + temp;
if((i+1)%2==0){
code += r.nextInt(10);
code += r.nextInt(10);
}
}
return code;
}
then i create on OBJ-C
-(void)generateCode{
NSString *alphabet = #"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXZY0123456789";
NSMutableString *s = [NSMutableString stringWithCapacity:4];
for (NSUInteger i = 0U; i < 4; i++) {
u_int32_t r = arc4random() % [alphabet length];
unichar c = [alphabet characterAtIndex:r];
[s appendFormat:#"%C", c];
}
NSLog(#"s-->%#",s);
}
but i got "HpNz" for result AC88pP13 insted that hve pattern String,string, numeric,numeric, lowescase string,numeric,numeric...
that case screw my life for 3 days...

Answer 1

Your Objective-C code looks good, but (as #Wain correctly said in a comment above),
the Java function function contains logic to insert 2 digits after 2 letters, which you
have not replicated in the Objective-C method.
I would make that logic slightly less obscure and write it as
- (void)generateCode
{
static NSString *letters = #"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXZY";
static NSString *digits = #"0123456789";
NSMutableString *s = [NSMutableString stringWithCapacity:8];
for (NSUInteger i = 0; i < 2; i++) {
uint32_t r;
// Append 2 random letters:
r = arc4random_uniform((uint32_t)[letters length]);
[s appendFormat:#"%C", [letters characterAtIndex:r]];
r = arc4random_uniform((uint32_t)[letters length]);
[s appendFormat:#"%C", [letters characterAtIndex:r]];
// Append 2 random digits:
r = arc4random_uniform((uint32_t)[digits length]);
[s appendFormat:#"%C", [digits characterAtIndex:r]];
r = arc4random_uniform((uint32_t)[digits length]);
[s appendFormat:#"%C", [digits characterAtIndex:r]];
}
NSLog(#"s-->%#",s);
}
Remark (from the man page):
arc4random_uniform(length) is preferred over arc4random() % length,
as it avoids "modulo bias" when the upper bound is not a power of two.
Remark: A more verbatim translation of the Java code code += r.nextInt(10);
to Objective-C would be
r = arc4random_uniform(10);
[s appendString:[#(r) stringValue]];
which creates a NSNumber object #(r) from the random number, and then
converts that to a string.

Answer 2

if you want a secure random string you should use this code:
#define ASCII_START_NUMERS 0x30
#define ASCII_END_NUMERS 0x39
#define ASCII_START_LETTERS_A 0x41
#define ASCII_END_LETTERS_Z 0x5A
#define ASCII_START_LETTERS_a 0x61
#define ASCII_END_LETTERS_z 0x5A
-(NSString *)getRandomString:(int)length {
NSMutableString *result = [[NSMutableString alloc]init];
while (result.length != length) {
NSMutableData* data = [NSMutableData dataWithLength:1];
SecRandomCopyBytes(kSecRandomDefault, 1, [data mutableBytes]);
Byte currentChar = 0;
[data getBytes:&currentChar length:1];
NSString *s = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
if (currentChar > ASCII_START_NUMERS && currentChar < ASCII_END_NUMERS) { // 0 to 0
[result appendString:s];
continue;
}
if (currentChar > ASCII_START_LETTERS_A && currentChar < ASCII_END_LETTERS_Z) { // A to Z
[result appendString:s];
continue;
}
if (currentChar > ASCII_START_LETTERS_a && currentChar < ASCII_END_LETTERS_z) { // a to z
[result appendString:s];
continue;
}
}
return result;
}