blocks|key|1397322|text|NSIntegers不是对象，您可以将它们转换为long，以便与当前64位体系结构的定义相匹配：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1397323|NSString+*inStr+=+[NSString+stringWithFormat:+@"%25ld",+(long)month];|1397324|entityMap^0|O|4|0|0|1V|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

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

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

blocks|key|1397362|text|答案已经给出，但我认为在某些情况下，这也是从NSInteger获取字符串的有趣方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1397363|NSInteger+value+=+12;
NSString+*+string+=+[NSString+stringWithFormat:@"%250.0f",+(float)value];|code-block|syntax|javascript|1397364|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

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

blocks|key|1181886|text|Obj-C+way+=)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1181887|NSString+*inStr+=+[@(month)+stringValue];|code-block|syntax|javascript|1181888|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Obj-C way =):

<pre><code>NSString *inStr = [@(month) stringValue];
</code></pre>

blocks|key|1181909|text|在支持arm64的情况下编译时，这不会生成警告：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1181910|[NSString+stringWithFormat:@"%25lu",+(unsigned+long)myNSUInteger];|code-block|syntax|javascript|1181911|entityMap^0|3|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

When compiling with support for <code>arm64</code>, this won't generate a warning:

<pre><code>[NSString stringWithFormat:@"%lu", (unsigned long)myNSUInteger];
</code></pre>

blocks|key|1181937|text|现代Objective-C|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1181938|NSInteger具有甚至可以与文字一起使用的stringValue方法|offset|length|style|CODE|1181939|NSString+*integerAsString1+=+[@12+stringValue];

NSInteger+number+=+13;
NSString+*integerAsString2+=+[@(number)+stringValue];|code-block|syntax|javascript|1181940|非常简单。难到不是么？|1181941|斯威夫特|1181942|var+integerAsString+=+String(integer)|1181943|entityMap^0|0|0|9|N|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@$D|W|E|X|F|G]|$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|J|7|13|8|@]|9|@]|A|$K|L]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

<h2>Modern Objective-C</h2>

An <code>NSInteger</code> has the method <code>stringValue</code> that can be used even with a literal

<pre><code>NSString *integerAsString1 = [@12 stringValue];

NSInteger number = 13;
NSString *integerAsString2 = [@(number) stringValue];
</code></pre>

Very simple. Isn't it?

<h2>Swift</h2>

<pre><code>var integerAsString = String(integer)
</code></pre>

blocks|key|1397439|text|您还可以尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1397440|NSInteger+month+=+1;
NSString+*inStr+=+[NSString+stringWithFormat:+@"%25ld",+month];|code-block|syntax|javascript|1397441|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can also try:

<pre><code>NSInteger month = 1;
NSString *inStr = [NSString stringWithFormat: @"%ld", month];
</code></pre>

blocks|key|1181983|text|在这种情况下，NSNumber可能对您有好处。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1181984|NSString+*inStr+=+[NSString+stringWithFormat:@"%25d",+
++++++++++++++++++++[NSNumber+numberWithInteger:[month+intValue]]];|code-block|syntax|javascript|1181985|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

NSNumber may be good for you in this case.

<pre><code>NSString *inStr = [NSString stringWithFormat:@"%d", 
 [NSNumber numberWithInteger:[month intValue]]];
</code></pre>

blocks|key|1182026|text|%25zd适用于NSIntegers+(适用于NSUInteger的%25tu)，在32位和64位架构上都没有强制转换和警告。我不知道为什么这不是"recommended+way“。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1182027|NSString+*string+=+[NSString+stringWithFormat:@"%25zd",+month];|code-block|syntax|javascript|1182028|如果您对这种方法的工作原理感兴趣，请参阅this+question。|1182029|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/library/mac/documentation/Cocoa/Conceptual/Strings/Articles/formatSpecifiers.html|1|https://stackoverflow.com/questions/18893880/alternatives-to-type-casting-when-formatting-nsuinteger-on-32-and-64-bit-archi^0|0|3|W|3|1Y|F|0|0|0|K|D|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@]|D|@$9|16|A|17|1|18]]|E|$]]|$1|M|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

<code>%zd</code> works for NSIntegers (<code>%tu</code> 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 "<a href="https://developer.apple.com/library/mac/documentation/Cocoa/Conceptual/Strings/Articles/formatSpecifiers.html" rel="nofollow noreferrer">recommended way</a>".

<pre><code>NSString *string = [NSString stringWithFormat:@"%zd", month];
</code></pre>

If you're interested in why this works see <a href="https://stackoverflow.com/questions/18893880/alternatives-to-type-casting-when-formatting-nsuinteger-on-32-and-64-bit-archi">this question</a>.

blocks|key|1182057|text|简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1182058|NSInteger+value+=+x;
NSString+*string+=+[@(value)+stringValue];|code-block|syntax|javascript|1182059|在这里，@(value)将给定的NSInteger转换为NSNumber对象，您可以为该对象调用所需的函数stringValue。|offset|length|style|CODE|1182060|entityMap^0|0|0|4|8|G|9|S|8|1H|B|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|N|$]]

Easy way to do:

<pre><code>NSInteger value = x;
NSString *string = [@(value) stringValue];
</code></pre>

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

How does one convert <code>NSInteger</code> to the <code>NSString</code> datatype?

I tried the following, where month is an <code>NSInteger</code>:

<pre><code> NSString *inStr = [NSString stringWithFormat:@"%d", [month intValue]];
</code></pre>

How do I convert NSInteger to NSString datatype?

Objective-C

如何将NSInteger转换为NSString数据类型？我尝试了以下方法，其中NSInteger是一个月  NSString *inStr = [NSString stringWithFormat:@"%d", [month intValue]];

问如何将NSInteger转换为NSString数据类型？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何将NSInteger转换为NSString数据类型？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何将NSInteger转换为NSString数据类型？
EN