blocks|key|2859597|text|属性b确实是一个可选的数组，但它的成员元素不是可选的。因此，当尝试将新值分配给现有成员时，新值必须是具体类型String，而不是可选类型String?。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2859598|//+this+is+an+Optional+array,+with+non-optional+elements
var+foo:+[String]?+=+["foo"]

let+optStr:+String?+=+"bar"
let+nonOptStr+=+"bar"

//+ok,+nonOptStr+is+not+an+optional
foo?[0]+=+nonOptStr

//+as+for+'nonOptStr',+we+need+to+unwrap
//+it+prior+to+attempting+to+assign+it+as+a+
//+replacement+member+of+'foo'
if+let+str+=+optStr+{
++++foo?[0]+=+str
}|code-block|syntax|javascript|2859599|此外，请避免显式展开，并考虑在使用索引之前对其进行验证。例如：|2859600|class+C+{
++++var+b:+[String]?
++++subscript(i:+Int)+->+String?+{
++++++++get+{
++++++++++++if+b?.indices.contains(i)+??+false+{+return+b?[i]+}
++++++++++++return+nil++++++++++++
++++++++}
++++++++set+{
++++++++++++if+b?.indices.contains(i)+??+false,+let+newValue+=+newValue+{
++++++++++++++++b?[i]+=+newValue
++++++++++++}
++++++++}
++++}
}|2859601|entityMap^0|2|1|1I|6|1W|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

You property <code>b</code> is indeed an optional array, but its member elements are not optionals. Hence, when attempting to assign a new value to an existing member, the new value must be of concrete type <code>String</code>, and not the optional type <code>String?</code>.

<pre><code>// this is an Optional array, with non-optional elements
var foo: [String]? = ["foo"]

let optStr: String? = "bar"
let nonOptStr = "bar"

// ok, nonOptStr is not an optional
foo?[0] = nonOptStr

// as for 'nonOptStr', we need to unwrap
// it prior to attempting to assign it as a 
// replacement member of 'foo'
if let str = optStr {
 foo?[0] = str
}
</code></pre>

Also, avoid explicit unwrapping, and consider validating the index prior to using it. E.g.:

<pre><code>class C {
 var b: [String]?
 subscript(i: Int) -&gt; String? {
 get {
 if b?.indices.contains(i) ?? false { return b?[i] }
 return nil 
 }
 set {
 if b?.indices.contains(i) ?? false, let newValue = newValue {
 b?[i] = newValue
 }
 }
 }
}
</code></pre>

blocks|key|2648380|text|你的变量b:字符串？是可选的数组。但是，其中的字符串不是可选的。如果您将其更改为:+var+b:+String?？然后它就会起作用了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2648381|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Your var b: [String]? is an optional array. However the string inside is not optional. If you changed it to: var b: [String?]? then it would work.

blocks|key|2648411|text|更普遍的是，我发现自己在对字符串进行拆分后出现了这个确切的错误，并最终打印出了这两个变量。这说明我试图将一个包含两个元素的变量赋值给一个包含字符串的索引数组。所以，这就是错误的来源，这实际上是说，您试图将一个数组的两个元素分配到另一个数组中的字符串中。好吧，这是有意义的错误。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2648412|算法是:+1)+array1+=+array2Item1，array2Item2|2648413|这就是为什么这很好:+2)+array1+=+array2Item1，array2Item2|2648414|但上述情况并非如此。|2648415|因此，正确的做法是2)，然后按索引分配：|2648416|array1+=+array2Item1，array2Item2|2648417|array3+=+array1去做我认为我在1中正在做的事情。|2648418|对于其他在这里冒险的人来说，这只是另一种解释。|2648419|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|U|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|V|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|W|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|X|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|Y|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|Q|$]]

More generally, I found myself with this exact error after doing a split on a string and ended up printing out both variables. That clarified that I was trying to assign a two element variable to an indexed array that contained a string. So, that is the source of the error, which is essentially saying your trying to assign two elements of an array into a string inside your other array. Well, that makes sense out of the error. 
Algorithmically it is:
1) array1[0] = array2Item1, array2Item2

That's why this was fine:
2) array1 = array2Item1, array2Item2

but the above was not.

So, the correct thing to do is 2) and then assign it by index:

array1 = array2Item1, array2Item2

array3 = array1[0] to do what I thought I was doing in 1).

Just another explanation for any others who venture here.

<pre><code>class C{
 var b:[String]?
 subscript(i:Int)-&gt;String?{
 get{
 return b?[i]
 }
 set{
 b?[i] = newValue! // notice the unwrapped “!” here
 }
 }
}
</code></pre>

In the code, I put a exclamation mark "!" to unwrap the <code>newValue</code>,
Because <code>newValue</code> is the same type of the return value of subscript which is String?.
If I don't put the exclamation mark "!" it will report error:
 "

<blockquote>
 error: cannot assign value of type 'String?' to type '<code>String</code>' <code>b?[i] = newValue</code>
</blockquote>

"
Question: <code>b?[i]</code> is obviously an optional String value String?, how come it is of type String.
I wonder it was a bad error code

Cannot assign value of type 'String?' to type 'String'

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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class C{ var b:[String]? subscript(i:Int)->String?{  get{   return b?[i]  }  set{   b?[i] = newValue! // notice the unwrapped “!” here  } }}在代码中，我放了一个感叹号"!“要解开n...

问不能赋值类型为'String?‘键入'String‘
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