blocks|key|1026204|text|NSNumber对象是不可变的；您可以做的最好的事情就是获取原始值，递增它，然后将结果包装在它自己的NSNumber对象中：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1026205|NSNumber+*bNumber+=+[NSNumber+numberWithInt:[aNumber+intValue]+%2B+1];|code-block|syntax|javascript|1026206|entityMap^0|0|8|1E|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

<code>NSNumber</code> objects are immutable; the best you can do is to grab the primitive value, increment it then wrap the result in its own <code>NSNumber</code> object:

<pre><code>NSNumber *bNumber = [NSNumber numberWithInt:[aNumber intValue] + 1];
</code></pre>

blocks|key|811545|text|对于使用最新版本的Xcode+(从4.4.1开始编写，SDK+5.1)的任何人来说，通过使用对象字面量，您可以将代码清理得更干净一点……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|811546|NSNumber+*x+=+@(1);
x+=+@([x+intValue]+%2B+1);
//+x+=+2|code-block|syntax|javascript|811547|为了做简单的操作，处理装箱和拆箱操作仍然有点痛苦，但它正在变得更好，至少更短。|811548|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

For anyone who is using the latest version of Xcode (writing this as of 4.4.1, SDK 5.1), with the use of object literals, you can clean the code up even a little bit more...

<pre><code>NSNumber *x = @(1);
x = @([x intValue] + 1);
// x = 2
</code></pre>

Still kind of a pain to deal with the boxing and unboxing everything to do simple operations, but it's getting better, or at least shorter.

blocks|key|811478|text|NSNumbers是不可变的，你必须创建一个新的实例。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|811479|//+Work+with+64+bit+to+support+very+large+values
myNSNumber+=+[NSNumber+numberWithLongLong:[myNSNumber+longLongValue]+%2B+1];

//+EDIT:+With+modern+syntax:
myNSNumber+=+@([myNSNumber+longLongValue]+%2B+1);|code-block|syntax|javascript|811480|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|$]]

NSNumbers are immutable, you have to create a new instance.

<pre><code>// Work with 64 bit to support very large values
myNSNumber = [NSNumber numberWithLongLong:[myNSNumber longLongValue] + 1];

// EDIT: With modern syntax:
myNSNumber = @([myNSNumber longLongValue] + 1);
</code></pre>

blocks|key|1026373|text|把它们放在一起，你就有了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1026374|myNSNumber+=+@(myNSNumber.intValue+%2B+1);|code-block|syntax|javascript|1026375|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|$]]

Put them all together, and you have:

<pre><code>myNSNumber = @(myNSNumber.intValue + 1);
</code></pre>

blocks|key|1026451|text|如果你用它来存储一个整数值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1026452|myNSNumber+=+@(myNSNumber.longLongValue+%2B+1);|code-block|syntax|javascript|1026453|对于浮点数，简短的答案如下，但这样做不是一个好主意，你会失去精确度，如果你稍后要做任何类型的比较，比如myNSNumber+is+isEqual:@(4.5)，你可能会感到惊讶：|1026454|myNSNumber+=+@(myNSNumber.floatValue+%2B+1);|1026455|如果您需要对Objective-C中表示为对象的浮点数进行数学运算(例如，如果您需要将它们放入数组、字典等中)您应该使用NSDecimalNumber。|offset|length|style|CODE|1026456|entityMap^0|0|0|0|0|1O|F|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

If you're using it to store an integral value:

<pre><code>myNSNumber = @(myNSNumber.longLongValue + 1);
</code></pre>

For floating point stuff the short answer is the following, but it's a bad idea to do this, you'll loose precision and if you're doing any kind of comparison later like [myNSNumber isEqual:@(4.5)] you might be surprised:

<pre><code>myNSNumber = @(myNSNumber.floatValue + 1);
</code></pre>

If you need to do math on floating point numbers represented as objects in Objective-C (i.e. if you need to put them in arrays, dictionaries, etc.) you should use <code>NSDecimalNumber</code>.

blocks|key|811670|text|使用类别可以简化将来的使用。这是一个基本的想法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|811671|+-+(void)incrementIntBy:(int)ammount+{
++++++self+=+[NSNumber+numberWithInt:(self.intValue+%2B+ammount)];
+}|code-block|syntax|javascript|811672|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|$]]

Use a category to ease future use. Here is a basic idea.

<pre><code> - (void)incrementIntBy:(int)ammount {
 self = [NSNumber numberWithInt:(self.intValue + ammount)];
 }
</code></pre>

blocks|key|811732|text|这个问题的答案中有很多很好的信息。我在我的代码中使用了选定的答案，它起作用了。然后我开始阅读其余的帖子，并大大简化了代码。如果您不熟悉Xcode5中引入的符号更改，阅读起来会有些困难，但它要清晰得多。我也许可以只写一行，但是要弄清楚我在做什么就有点太难了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|811733|我在字典中存储了一个NSNumber，并且我想要递增它。我得到它，将它转换为一个整数，在将它转换为NSNumber时递增它，然后将它放回字典中。|811734|旧代码|811735|+NSNumber+*correct+=+[scoringDictionary+objectForKey:@"correct"];
+int+correctValue+=+[correct+intValue];
+correct+=+[NSNumber+numberWithInt:correctValue+%2B+1];
+[scoringDictionary+removeObjectForKey:@"correct"];
+scoringDictionary[@"correct"]+=+correct;|code-block|syntax|javascript|811736|新代码|811737|NSNumber+*correct+=+[scoringDictionary+objectForKey:@"correct"];
scoringDictionary[@"correct"]+=+@(correct.intValue+%2B+1);|811738|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Lots of good info in the answers to this question. I used the selected answer in my code and it worked. Then I started reading the rest of the posts and simplified the code a lot. It’s a bit harder to read if you aren’t familiar with the changes in notation that were introduced in Xcode 5, but it’s a lot cleaner. I probably could make it just one line, but then it’s a little too hard to figure out what I’m doing.

I’m storing an NSNumber in a dictionary and I want to increment it. I get it, convert it to an int, increment it while converting it to an NSNumber, then put it back into the dictionary.

Old Code

<pre><code> NSNumber *correct = [scoringDictionary objectForKey:@"correct"];
 int correctValue = [correct intValue];
 correct = [NSNumber numberWithInt:correctValue + 1];
 [scoringDictionary removeObjectForKey:@"correct"];
 scoringDictionary[@"correct"] = correct;
</code></pre>

New code

<pre><code>NSNumber *correct = [scoringDictionary objectForKey:@"correct"];
scoringDictionary[@"correct"] = @(correct.intValue + 1);
</code></pre>

问如何递增NSNumber
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