ios 如何将 NSInteger 转换为 NSString 数据类型？
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原文地址: http://stackoverflow.com/questions/1796390/
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How do I convert NSInteger to NSString datatype?
提问by senthilMuthu
How does one convert NSIntegerto the NSStringdatatype?
如何转换NSInteger为NSString数据类型？
I tried the following, where month is an NSInteger:
我尝试了以下操作，其中月份是一个NSInteger：
NSString *inStr = [NSString stringWithFormat:@"%d", [month intValue]];
回答by luvieere
NSIntegers are not objects, you cast them to long, in order to match the current 64-bit architectures' definition:
NSIntegers 不是对象，您将它们强制转换为long，以匹配当前 64 位架构的定义：
NSString *inStr = [NSString stringWithFormat: @"%ld", (long)month];
NSString *inStr = [NSString stringWithFormat: @"%ld", (long)month];
回答by Alexey Kozhevnikov
Obj-C way =):
Obj-C 方式 =)：
NSString *inStr = [@(month) stringValue];
回答by MadNik
Modern Objective-C
现代Objective-C
An NSIntegerhas the method stringValuethat can be used even with a literal
AnNSInteger具有stringValue即使使用文字也可以使用的方法
NSString *integerAsString1 = [@12 stringValue];
NSInteger number = 13;
NSString *integerAsString2 = [@(number) stringValue];
Very simple. Isn't it?
很简单。不是吗？
Swift
迅速
var integerAsString = String(integer)
回答by Kevin
%zdworks for NSIntegers (%tufor 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".
%zd适用于 NSIntegers（%tu对于 NSUInteger），在 32 位和 64 位架构上没有强制转换和警告。我不知道为什么这不是“推荐的方式”。
NSString *string = [NSString stringWithFormat:@"%zd", month];
If you're interested in why this works see this question.
如果您对为什么这样做感兴趣，请参阅此问题。
回答by Karthik damodara
Easy way to do:
简单的做法：
NSInteger value = x;
NSString *string = [@(value) stringValue];
Here the @(value)converts the given NSIntegerto an NSNumberobject for which you can call the required function, stringValue.
这里@(value)将给定的转换NSInteger为一个NSNumber对象，您可以为其调用所需的函数stringValue.
回答by Andreas Ley
When compiling with support for arm64, this won't generate a warning:
编译时支持arm64，这不会产生警告：
[NSString stringWithFormat:@"%lu", (unsigned long)myNSUInteger];
回答by NeverHopeless
You can also try:
你也可以试试：
NSInteger month = 1;
NSString *inStr = [NSString stringWithFormat: @"%ld", month];
回答by hothead
NSNumber may be good for you in this case.
在这种情况下， NSNumber 可能对你有好处。
NSString *inStr = [NSString stringWithFormat:@"%d",
[NSNumber numberWithInteger:[month intValue]]];
回答by Nazir
The answer is given but think that for some situation this will be also interesting way to get string from NSInteger
给出了答案，但认为在某些情况下，这也是从 NSInteger 获取字符串的有趣方式
NSInteger value = 12;
NSString * string = [NSString stringWithFormat:@"%0.0f", (float)value];
