NSInteger到NSString误解

Question

I have a question regarding NSInteger to NSString i keep trying but with no luck. however for some strange reason one of my NSInteger does but i dont know why i have this method:

//Lets say loc_id = 869
-(void)addResource:(NSInteger)loc_id
{
   //[loc.chosenLocations addObject:[NSString stringWithFormat@"%@",loc_id]];
   NSLog([NSString stringWithFormat:@"addResource: loc_id:%@",loc_id]);
}

Results BAD

Because its giving me: addResource: loc_id:20987787696 in the debugger of xcode

Its ment to be loc_id:860

when i try this with loc_id = 8 it works pretty well why is this strange behaviour or what did i mis understood thanks for the tips and help. when i use %@ in string format it works for loc_id = 8 but when i try to use %d wich is normal useage it wil also result in loc_id:20987787696 how come ?

Useage of method:

-(IBAction)switchChanged:(id)sender {
  UISwitch *switchControl = sender;
  [loc addResource:switchControl.tag] 
 }

 -(void)addSwitch:(NSInteger)aid
 {
    UISwitch *sw = [[UISwitch alloc]initWithFrame:CGRectMake(x,y,w,h)]];
    sw.tag = aid;
    [sw addTarget:self action:@selector(switchChanged:) forControlEvents:UIControlEvenValueChanged]];
    [self.View addSubview:sw];
 }

 //Data is dynamic (NSObject) but for example i will exclude the dynamic data.
 for(int i = 0; i < 20; i++)
 {
   NSInteger *loc_id = 868;
   [self addSwitch:loc_id];

 }

at a point i want to insert it into the NSMutableArray as commented out above but i expect a value of 869 and it giefs me 2071894069 i want the value of the int not the memory thingy

Kind Regards, Stefan

Answer 1

You should use %li or @ld instead of %@ , and drop the stringWithFormat: call in the NSLog (it already expects a format):

NSLog(@"addResource: loc_id:%li",loc_id);

%@ is the selector to format objects descriptions and strings. %li is to cast your integer to a long, and prevent an eventual compiler warning.

---

Edit :

After quite a few questions in the comment, here is another wrong thing in your code :

NSInteger *loc_id = 868;

You should not initiate loc_id as a pointer. Use instead :

NSInteger loc_id = 868;

---

Edit #2 :

As discussed in the chat, the issue came from the way the NSInteger value was extracted from the array / dictionary in which it was held.

Answer 2

To display your NSInteger, you don't need to format it in NSString. Just use this code below.

NSLog(@" %ld",(long)loc_id);

Answer 3

You need to use %ld or %li for integers. %@ is used for objects.

NSLog([NSString stringWithFormat:@"addResource:loc_id:%ld",(long)loc_id]);

Answer 4

You also can cast your Integer to a long value:

-(void)addResource:(NSInteger)loc_id
{
    NSLog(@"addResource: loc_id:%ld",(long)loc_id);
}

Answer 5

Apple Documentation

The utility functions NSLog() and NSLogv() use the NSString string formatting services to log error messages. Note that as a consequence of this, you should take care when specifying the argument for these functions. A common mistake is to specify a string that includes formatting characters, as shown in the following example.

Wrong

You don't need to add formatting strings inside NSLog:

NSLog([NSString stringWithFormat:@"addResource: loc_id:%@",loc_id]);

Right

Also using %li on 64 Bit gives integer precision as @rdurand asnwer

NSLog(@"addResource: loc_id:%li",loc_id);

Answer 6

    NSLog([NSString stringWithFormat:@"addResource: loc_id:%ld",(long)loc_id]);

Gets a warning: Format string is not a string literal.

The following code works as ordered:

-(void)addResource:(NSInteger)loc_id
{
    //[loc.chosenLocations addObject:[NSString stringWithFormat@"%@",loc_id]];
    NSString *string = [NSString stringWithFormat:@"addResource: loc_id:%ld",(long)loc_id];
    NSLog(@"%@", string);
}