Question

I’m new in Objective-C programmation and I receive this error when I run my little program (I just want to generate a random char). So here’s what I’ve done so far:

(IBAction)generate{
  int a = arc4random() % 26;
  NSString * chaine = @"ABCDEFGHIJKLMNOPQRSTUVWXYZ";
  char lettre = [chaine characterAtIndex:a];

  NSMutableString * mot = [[NSMutableString alloc] initWithCharacters:lettre length:1];
  hasard.text = mot;
}

I tried to simply put my variable ‘lettre’ in hasard.text but it won’t work with the error ‘Incompatible integer to pointer conversion assigning to ‘NSString *‘ from ‘char‘. So I created an NSMutableString to contain my character.

When I put the character “e” manually instead of the variable ‘letter’ on the fifth line it works well. Since I can see in the debugger that ‘lettre’ contains a random letter, why do I get the error in the title ?

(EXC_BAD_ACCESS (code=2, address=0x42)).

Answer 1

The function is expecting a pointer to a character, you are giving it a literal character. You need to create a pointer to it. Also, characterAtIndex doesn’t return a character as you would think. It returns a unicode character which is actually an unsigned short instead of an unsigned char. However, if you change your code to this it will work:

const unichar foo = [@"Test" characterAtIndex:0];
NSString *test = [NSString stringWithCharacters:&foo length:1]; //Note the &

EDIT But the simplest way would just be this:

char randChar = arc4random_uniform(26) + 'A'; //Changed in response to Jason Coco's comment
NSString *mot = [NSString stringWithFormat:@"%c", randChar];