How do I convert NSInteger to NSString datatype?

Question

How does one convert NSInteger to the NSString datatype?

I tried the following, where month is an NSInteger:

  NSString *inStr = [NSString stringWithFormat:@"%d", [month intValue]];

Answer 1

NSIntegers are not objects, you cast them to long, in order to match the current 64-bit architectures' definition:

NSString *inStr = [NSString stringWithFormat: @"%ld", (long)month];

Answer 2

Obj-C way =):

NSString *inStr = [@(month) stringValue];

Answer 3

Modern Objective-C

An NSInteger has the method stringValue that can be used even with a literal

NSString *integerAsString1 = [@12 stringValue];

NSInteger number = 13;
NSString *integerAsString2 = [@(number) stringValue];

Very simple. Isn't it?

Swift

var integerAsString = String(integer)

Answer 4

%zd works for NSIntegers (%tu for NSUInteger) with no casts and no warnings on both 32-bit and 64-bit architectures. I have no idea why this is not the "recommended way".

NSString *string = [NSString stringWithFormat:@"%zd", month];

If you're interested in why this works see this question.

Answer 5

Easy way to do:

NSInteger value = x;
NSString *string = [@(value) stringValue];

Here the @(value) converts the given NSInteger to an NSNumber object for which you can call the required function, stringValue.

Answer 6

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

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

Answer 7

You can also try:

NSInteger month = 1;
NSString *inStr = [NSString stringWithFormat: @"%ld", month];

Answer 8

The answer is given but think that for some situation this will be also interesting way to get string from NSInteger

NSInteger value = 12;
NSString * string = [NSString stringWithFormat:@"%0.0f", (float)value];

Answer 9

NSNumber may be good for you in this case.

NSString *inStr = [NSString stringWithFormat:@"%d", 
                    [NSNumber numberWithInteger:[month intValue]]];