Question

While deploying my application, I got the error message: “Thread 1:Program received signal: “EXC_BAD_ACCESS”.

My code is below:

-(NSDictionary *)syncWithList:(NSInteger)listID
{
    NSString *urlit = [NSString stringWithFormat:@"http://0.0.0.0:3000/lists/%@/syncList.json?auth_token=%@",@"xxxxxxxxxxx",listID];
// **Here I got the error message: "Thread 1:Program received signal: "EXC_BAD_ACCESS"**
    NSLog(@"url: %@",urlit);
    NSURL *freequestionurl = [NSURL URLWithString:urlit];
    ASIHTTPRequest *back = [ASIHTTPRequest requestWithURL:freequestionurl];
    [back startSynchronous];
    self.listData = [[back responseString] objectFromJSONString];
    NSLog(@"%@",listData);
    NSDictionary *dicPost = [listData objectAtIndex:0];
    return dicPost;
}

Thanks a lot!!!!

Answer 1

You must not format NSInteger (which is just a typedef’d int on current iOS versions) with the %@ specifier. Writing %@ in a string format basically means “call description on the object and use the result”.
But NSInteger is not an object, it’s a primitive type.
You get a memory exception because when listID is 42 you access an object at memory address 42. This is definitely not what you want.

-(NSDictionary *)syncWithList:(NSInteger)listID
                               ^^^^^^^^^
NSString *urlit = [NSString stringWithFormat:@"http://0.0.0.0:3000/lists/%@/syncList.json?auth_token=%@",@"xxxxxxxxxxx",listID];
                                                                                                     ^^

just use the %i format specifier instead of %@ for listID.

NSString *urlit = [NSString stringWithFormat:@"http://0.0.0.0:3000/lists/%@/syncList.json?auth_token=%i",@"xxxxxxxxxxx",listID];