How should I pass an int into stringWithFormat?

Objective CCocoa Touch

Objective C Problem Overview


I am try to use stringWithFormat to set a numerical value on the text property of a label but the following code is not working. I cannot cast the int to NSString. I was expecting that the method would know how to automatically convert an int to NSString.

What do I need to do here?

- (IBAction) increment: (id) sender
{
	int count = 1;
	label.text = [NSString stringWithFormat:@"%@", count];
}

Objective C Solutions


Solution 1 - Objective C

Do this:

label.text = [NSString stringWithFormat:@"%d", count];

Solution 2 - Objective C

Keep in mind that @"%d" will only work on 32 bit. Once you start using NSInteger for compatibility if you ever compile for a 64 bit platform, you should use @"%ld" as your format specifier.

Solution 3 - Objective C

Marc Charbonneau wrote: > Keep in mind that @"%d" will only work on 32 bit. Once you start using NSInteger for compatibility if you ever compile for a 64 bit platform, you should use @"%ld" as your format specifier.

Interesting, thanks for the tip, I was using @"%d" with my NSIntegers!

The SDK documentation also recommends to cast NSInteger to long in this case (to match the @"%ld"), e.g.:

NSInteger i = 42;
label.text = [NSString stringWithFormat:@"%ld", (long)i];

Source: String Programming Guide for Cocoa - String Format Specifiers (Requires iPhone developer registration)

Solution 4 - Objective C

You want to use %d or %i for integers. %@ is used for objects.

It's worth noting, though, that the following code will accomplish the same task and is much clearer.

label.intValue = count;

Solution 5 - Objective C

And for comedic value:

label.text = [NSString stringWithFormat:@"%@", [NSNumber numberWithInt:count]];

(Though it could be useful if one day you're dealing with NSNumber's)

Solution 6 - Objective C

To be 32-bit and 64-bit safe, use one of the Boxed Expressions:

  label.text = [NSString stringWithFormat:@"%@", @(count).stringValue];

Solution 7 - Objective C

Is the snippet you posted just a sample to show what you are trying to do?

The reason I ask is that you've named a method increment, but you seem to be using that to set the value of a text label, rather than incrementing a value.

If you are trying to do something more complicated - such as setting an integer value and having the label display this value, you could consider using bindings. e.g

You declare a property count and your increment action sets this value to whatever, and then in IB, you bind the label's text to the value of count. As long as you follow Key Value Coding (KVC) with count, you don't have to write any code to update the label's display. And from a design perspective you've got looser coupling.

Solution 8 - Objective C

Don't forget for long long int:

long long int id = [obj.id longLongValue];
[NSString stringWithFormat:@"this is my id: %lld", id]

Solution 9 - Objective C

NSString * formattedname;
NSString * firstname;
NSString * middlename;
NSString * lastname;

firstname = @"My First Name";
middlename = @"My Middle Name";
lastname = @"My Last Name";

formattedname = [NSString stringWithFormat:@"My Full Name: %@ %@ %@", firstname, middlename, lastname];
NSLog(@"\n\nHere is the Formatted Name:\n%@\n\n", formattedname);

/*
Result:
Here is the Formatted Name:
My Full Name: My First Name My Middle Name My Last Name
*/

Solution 10 - Objective C

label.text = [NSString stringWithFormat:@"%d", XYZ]; 

//result:   label.text = XYZ
//use %d for int values

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QuestionBrennanView Question on Stackoverflow
Solution 1 - Objective CBobbyShaftoeView Answer on Stackoverflow
Solution 2 - Objective CMarc CharbonneauView Answer on Stackoverflow
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Solution 4 - Objective CZach LangleyView Answer on Stackoverflow
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