How do you convert an NSUInteger into an NSString?

Question

How do you convert an NSUInteger into an NSString? I've tried but my NSString returned 0 all the time.

NSUInteger NamesCategoriesNSArrayCount = [self.NamesCategoriesNSArray count];  
NSLog(@"--- %d", NamesCategoriesNSArrayCount);  
[NamesCategoriesNSArrayCountString setText:[NSString stringWithFormat:@"%d",    NamesCategoriesNSArrayCount]];  
NSLog(@"=== %d", NamesCategoriesNSArrayCountString);

Simulcast

Answer 1

When compiling with support for arm64, this won't generate a warning:

[NSString stringWithFormat:@"%lu", (unsigned long)myNSUInteger];

Orling

or %zd for NSUInteger, without typecasting that is. – Pneumatograph

@PaulPeelen Did you mean to say NSInteger? – Orling

My bad.. I meant, of course, %tu (since its unsigned). – Pneumatograph

Answer 2

I hope your NamesCategoriesNSArrayCountString is NSString; if yes use the below line of code.

NamesCategoriesNSArrayCountString  = [NSString stringWithFormat:@"%d", NamesCategoriesNSArrayCount]];

istead of

[NamesCategoriesNSArrayCountString setText:[NSString stringWithFormat:@"%d", NamesCategoriesNSArrayCount]];

Masturbation

As of Xcode 4.5, this will throw a "Data argument not used by format string" warning. Does anyone know the correct format? – Zajac

Disregard last. I had a typo that I didn't notice for the longest time, and the typo was what was causing the error. [NSString stringWithFormat:@"%d", myNSUInteger] is still the correct way to do this. – Zajac

The answer by @andreas-ley is more accurate. %d is the format string for a signed integer. NSUInteger is an unsigned long integer. Using %lu is needed once the overlap between the unsigned and signed range is exceeded, otherwise a negative value will be printed when in fact the number is positive. – Sulphone

Answer 3

When compiling for arm64, use the following to avoid warnings:

[NSString stringWithFormat:@"%tu", myNSUInteger];

Or, in your case:

NSUInteger namesCategoriesNSArrayCount = [self.NamesCategoriesNSArray count];  
NSLog(@"--- %tu", namesCategoriesNSArrayCount);  
[namesCategoriesNSArrayCountString setText:[NSString stringWithFormat:@"%tu", namesCategoriesNSArrayCount]];  
NSLog(@"=== %@", namesCategoriesNSArrayCountString);

(Also, tip: Variables start with lowercase. Info: here)

Pneumatograph

Note that while tu works right now, it isn't guaranteed to. See here: #18894380 – Orling

Very true. I didn't know that actually. Good to keep in mind for the future. – Pneumatograph

Answer 4

You can also use:

NSString *rowString = [NSString stringWithFormat: @"%@",  @(row)];

where row is a NSUInteger.

Averroes

Answer 5

This String Format Specifiers article from Apple is specific when you need to format Apple types:

OS X uses several data types—NSInteger, NSUInteger,CGFloat, and CFIndex—to provide a consistent means of representing values in 32- and 64-bit environments. In a 32-bit environment, NSInteger and NSUInteger are defined as int and unsigned int, respectively. In 64-bit environments, NSInteger and NSUInteger are defined as long and unsigned long, respectively. To avoid the need to use different printf-style type specifiers depending on the platform, you can use the specifiers shown in Table 3. Note that in some cases you may have to cast the value.

• [NSString stringWithFormat:@"%ld", (long)value] : NSInteger displayed as decimal
• [NSString stringWithFormat:@"%lx", (long)value] : NSInteger displayed as hex
• [NSString stringWithFormat:@"%lu", (unsigned long)value] : NSUInteger displayed as decimal
• [NSString stringWithFormat:@"%lx", (unsigned long)value] : NSUInteger displayed as hex
• [NSString stringWithFormat:@"%f", value] : CGFloat
• [NSString stringWithFormat:@"%ld", (long)value] : CFIndex displayed as decimal
• [NSString stringWithFormat:@"%lx", (long)value] : CFIndex displayed as hex

See the article for more details.

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