blocks|key|4119997|text|给定的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4119998|enum+Foo+{Bar=0,Baz,...,Last};|code-block|syntax|javascript|4119999|您可以像这样迭代元素：|4120000|for(int+i=Bar;+i<=Last;+i%2B%2B)+{
++...
}|4120001|请注意，这暴露了C枚举的just-an-int性质。特别是，您可以看到C枚举并不真正提供类型安全，因为您可以使用int代替枚举值，反之亦然。此外，这取决于枚举值的声明顺序:至少可以说是脆弱的。此外，请参阅Chuck的注释；如果枚举项目是不连续的(例如，因为您为某些项目指定了显式的、非连续的值)，这将根本不起作用。啊呀。|4120002|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Given

<pre><code>enum Foo {Bar=0,Baz,...,Last};
</code></pre>

you can iterate the elements like:

<pre><code>for(int i=Bar; i&lt;=Last; i++) {
 ...
}
</code></pre>

Note that this exposes the really-just-an-int nature of a C enum. In particular, you can see that a C enum doesn't really provide type safety, as you can use an int in place of an enum value and vice versa. In addition, this depends on the declaration order of the enum values: fragile to say the least. In addition, please see Chuck's comment; if the enum items are non-contiguous (e.g. because you specified explicit, non-sequential values for some items), this won't work at all. Yikes.

blocks|key|4119938|text|如果您的枚举定义如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4119939|enum+Direction+{++East,++West,++North,++South};|code-block|syntax|javascript|4119940|你可以这样遍历它：|4119941|for+(+int+direction+=+East;+direction+<=+South;+%2B%2Bdirection)
{
+++/*+Do+something+with+Direction
}|4119942|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you enum is defined as follows:

<pre><code>enum Direction { East, West, North, South};
</code></pre>

You can loop through it this way:

<pre><code>for ( int direction = East; direction &lt;= South; ++direction)
{
 /* Do something with Direction
}
</code></pre>

blocks|key|4120015|text|这有点杂乱无章(就像整个概念一样……嗯……)但是如果你要多次这样做，并且你希望能够处理不连续枚举的可能性，你可以创建一个(全局常量)数组，并(使用ennukiller的例子)方向directions4+={东，西，北，南}；然后在你的循环中，你可以谈论directionsi，而不是直接迭代方向本身……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4120016|就像我说的，它很丑陋，但我认为它不那么脆弱……|4120017|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

It's a bit of a kludge (as is the entire concept so... eh...) but if you're going to do this multiple times, and you want to be able to deal with the possibility of non-contiguous enums, you can create a (global constant) array, and have (to use ennukiller's example) Directions directions[4] = {East, West, North, South};
and then in your loop you can talk about directions[i] rather than iterating directly over the directions themselves... 

As I said, it's ugly, but it's slightly less fragile I think...

blocks|key|1625075|text|对于不连续的枚举，您可以通过执行以下操作来迭代有效值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1625076|enum+{
++++value_one+++=+0,
++++value_two+++=+1,
++++value_three+=+3,
++++value_max
}
for+(+i+=+value_one;+i+<+value_max;+i%2B%2B+)
{+
++++switch(i)
++++{
++++++++case+value_one:
++++++++case+value_two:
++++++++case+value_three:
++++++++case+value_four:
++++++++{
++++++++++++/*+...+Do+something+with+valid+enum+value+...+*/
++++++++}
++++++++break;
++++++++default:
++++++++++++/*+Found+an+invalid+enum+entry;+ignore+it+*/
++++++++break;
++++}
}|code-block|syntax|javascript|1625077|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|$]]

For non-contiguous enums you can iterate over valid values by doing:

<pre><code>enum {
 value_one = 0,
 value_two = 1,
 value_three = 3,
 value_max
}
for ( i = value_one; i &lt; value_max; i++ )
{ 
 switch(i)
 {
 case value_one:
 case value_two:
 case value_three:
 case value_four:
 {
 /* ... Do something with valid enum value ... */
 }
 break;
 default:
 /* Found an invalid enum entry; ignore it */
 break;
 }
}
</code></pre>

blocks|key|1625065|text|为此，我编写了以下解决方案，满足了我自己的需求。我想我应该把它贴在这里，供人们搜索类似的东西。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1625066|假设我有一个枚举。|1625067|typedef+NS_ENUM+(NSInteger,+ISDirection)+{
++ISDirectionUp,
++ISDirectionDown,
++ISDirectionLeft,
++ISDirectionRight
};|code-block|syntax|javascript|1625068|为了能够遍历它，我定义了以下预处理器宏。|1625069|#define+ISDirectionAllAsDir+ISDirection+dir+=+ISDirectionUp;+dir+<=+ISDirectionRight;+dir%2B%2B|1625070|然后我在一个循环中使用它，如下所示：|1625071|for(ISDirectionAllAsDir)+{
++NSLog(@"Direction+%25d",+dir);
}|1625072|这将迭代逻辑放在一个地方(宏)，这增加了代码的可维护性。|1625073|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

In search of this, I just concocted the following solution which fulfilled my own requirements. I thought I'd post it here for people searching something similar.

Say I have an enum.

<pre><code>typedef NS_ENUM (NSInteger, ISDirection) {
 ISDirectionUp,
 ISDirectionDown,
 ISDirectionLeft,
 ISDirectionRight
};
</code></pre>

To be able to loop over it, I define the following preprocessor macro.

<pre><code>#define ISDirectionAllAsDir ISDirection dir = ISDirectionUp; dir &lt;= ISDirectionRight; dir++
</code></pre>

Then I use it inside a loop, as such:

<pre><code>for(ISDirectionAllAsDir) {
 NSLog(@"Direction %d", dir);
}
</code></pre>

This puts the logic for iterating in one place (the macro) which increases code maintainability.

blocks|key|4120121|text|任何枚举的长度(项数)都可以通过获取枚举中的名称数组并利用数组的长度来确定。使用枚举的GetNames方法。Enum.GetNames(enumType).Length.|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4120122|{
enum+Pig+{+Snort,+Sty,+Oink,+Curly+};
const+int+Pig_Count+=+Enum.GetNames(typeof(Pig)).Length;
}|code-block|syntax|javascript|4120123|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 length (count of items) of any enumeration can be ascertained by getting the array of names in the enumeration and utilizing the array's length. Use the GetNames method of Enum. Enum.GetNames(enumType).Length.

<pre><code>{
enum Pig { Snort, Sty, Oink, Curly };
const int Pig_Count = Enum.GetNames(typeof(Pig)).Length;
}
</code></pre>

blocks|key|2009629|text|我喜欢我从这里得到的Iteration+over+enum+in+Objective+C?|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2009630|typedef+enum+{+ENUM1=0,++ENUM2,+ENUM3,+ENUM4+,+TOTAL++}+;


for+(+int+i=ENUM1+;+i<TOTAL;+i%2B%2B)
{}|code-block|syntax|javascript|2009631|这样你以后就可以继续添加枚举，而不会影响你的循环？|2009632|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/6910127/iteration-over-enum-in-objective-c^0|A|Z|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

I like what I got from here
<a href="https://stackoverflow.com/questions/6910127/iteration-over-enum-in-objective-c">Iteration over enum in Objective C?</a>

<pre><code>typedef enum { ENUM1=0, ENUM2, ENUM3, ENUM4 , TOTAL } ;


for ( int i=ENUM1 ; i&lt;TOTAL; i++)
{}
</code></pre>

that way you can keep adding enums later on and not affect your loops?

blocks|key|4120161|text|如果由于您使用的是C%2B%2B而导致上述方法不起作用，那么可以使用the+following+should|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4120162|4120163|枚举Foo+{+One，Two，Three，Last+}；for+(+int+fooInt+=+One；fooInt+!=+Last；fooInt%2B%2B+){+foo+Foo+=+static_cast(fooInt)；//+...+}|blockquote|4120164|4120165|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/261986/1587329^0|U|K|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|C|$]]|$1|E|3|F|5|G|7|V|8|@]|9|@]|C|$]]|$1|H|3|-4|5|6|7|W|8|@]|9|@]|C|$]]|$1|I|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

If the above does not work for you, because you're using C++, <a href="https://stackoverflow.com/a/261986/1587329">the following should</a>:

<blockquote>
<pre><code>enum Foo {
 One,
 Two,
 Three,
 Last
};

for ( int fooInt = One; fooInt != Last; fooInt++ )
{
 Foo foo = static_cast&lt;Foo&gt;(fooInt);
 // ...
}
</code></pre>
</blockquote>

blocks|key|4120145|text|我讨厌用"+last“条目弄脏我的枚举(以及附带的说明，它不应该被使用，而且它总是必须保存在最后)，并且只找到了一种方法来绕过它(除了将最后一个枚举条目用作秘密的"last”，这再次让我感到讨厌)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4120146|如果使用X+Macros定义枚举，则可以非常简单地获得枚举中的项数。当您可能使用此方法时，请参阅FMI链接-为了使内容简短，请看下面的内容：|offset|length|4120147|#define+COLOR_TABLE+\
X(red,+"red")+++++++\
X(green,+"green")+++\
X(blue,+"blue")

#define+X(a,+b)+a,
enum+COLOR+{
++COLOR_TABLE
};
#undef+X

#define+X(a,+b)+b,
char+*color_name[]+=+{
++COLOR_TABLE
};
#undef+X

#define+X(a,+b)+%2B1
int+color_count+=+0+COLOR_TABLE;
#undef+X|code-block|syntax|javascript|4120148|entityMap|0|LINK|mutability|MUTABLE|url|http://www.drdobbs.com/the-new-c-x-macros/184401387^0|0|4|8|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@$D|U|E|V|1|W]]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

I hate dirtying up my enum with a "Last" entry (and the accompanying note that it shouldn't be used, and that it always hast to be kept, well, last), and have only found one way around it (other than using the last enum entry as a secret "last," which, again, makes me feel icky).

If you define your enum using <a href="http://www.drdobbs.com/the-new-c-x-macros/184401387" rel="nofollow noreferrer">X Macros</a>, you can get a count of the number of items in your enum pretty simply. See the link FMI on when you might use this approach - to keep things brief, here's what it looks like:

<pre class="lang-c prettyprint-override"><code>#define COLOR_TABLE \
X(red, "red") \
X(green, "green") \
X(blue, "blue")

#define X(a, b) a,
enum COLOR {
 COLOR_TABLE
};
#undef X

#define X(a, b) b,
char *color_name[] = {
 COLOR_TABLE
};
#undef X

#define X(a, b) +1
int color_count = 0 COLOR_TABLE;
#undef X
</code></pre>

Is it possible to loop through enum values in Objective-C?

looping through enum values

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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是否可以遍历Objective-C中的枚举值？

问遍历枚举值
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