问题
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]];
回答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];
回答2:
Obj-C way =):
NSString *inStr = [@(month) stringValue];
回答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)
回答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.
回答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.
回答6:
When compiling with support for arm64, this won't generate a warning:
[NSString stringWithFormat:@"%lu", (unsigned long)myNSUInteger];
回答7:
You can also try:
NSInteger month = 1;
NSString *inStr = [NSString stringWithFormat: @"%ld", month];
回答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];
回答9:
NSNumber may be good for you in this case.
NSString *inStr = [NSString stringWithFormat:@"%d",
[NSNumber numberWithInteger:[month intValue]]];
来源:https://stackoverflow.com/questions/1796390/how-do-i-convert-nsinteger-to-nsstring-datatype