blocks|key|705362|text|如下所示的扩展方法可以完成这项工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|705363|public+static+int+Your_Name_Here(this+string+s,+string+other)+
{
++++string+first+=+s.Length+<+other.Length+?+s+:+other;
++++string+second+=+s.Length+>+other.Length+?+s+:+other;

++++for+(int+counter+=+0;+counter+<+first.Length;+counter%2B%2B)
++++{
++++++++if+(first[counter]+!=+second[counter])
++++++++{
++++++++++++return+counter;
++++++++}
++++}
++++return+-1;
}|code-block|syntax|javascript|705364|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|$]]

An extension method along the lines of the below would do the job:

<pre><code>public static int Your_Name_Here(this string s, string other) 
{
 string first = s.Length &lt; other.Length ? s : other;
 string second = s.Length &gt; other.Length ? s : other;

 for (int counter = 0; counter &lt; first.Length; counter++)
 {
 if (first[counter] != second[counter])
 {
 return counter;
 }
 }
 return -1;
}
</code></pre>

blocks|key|489697|text|///+<summary>
///+Gets+a+first+different+char+occurence+index
///+</summary>
///+<param+name="a">First+string</param>
///+<param+name="b">Second+string</param>
///+<param+name="handleLengthDifference">
///+If+true+will+return+index+of+first+occurence+even+strings+are+of+different+length
///+and+same-length+parts+are+equals+otherwise+-1
///+</param>
///+<returns>
///+Returns+first+difference+index+or+-1+if+no+difference+is+found
///+</returns>
public+int+GetFirstBreakIndex(string+a,+string+b,+bool+handleLengthDifference)
{
++++int+equalsReturnCode+=+-1;
++++if+(String.IsNullOrEmpty(a)+%7C%7C+String.IsNullOrEmpty(b))
++++{
++++++++return+handleLengthDifference+?+0+:+equalsReturnCode;
++++}

++++string+longest+=+b.Length+>+a.Length+?+b+:+a;
++++string+shorten+=+b.Length+>+a.Length+?+a+:+b;++++
++++for+(int+i+=+0;+i+<+shorten.Length;+i%2B%2B)
++++{
++++++++if+(shorten[i]+!=+longest[i])
++++++++{
++++++++++++return+i;
++++++++}
++++}

++++//+Handles+cases+when+length+is+different+(a="1234",+b="123")
++++//+index=3+would+be+returned+for+this+case
++++//+If+you+do+not+need+such+behaviour+-+just+remove+this
++++if+(handleLengthDifference+&&+a.Length+!=+b.Length)
++++{
++++++++return+shorten.Length;
++++}

++++return+equalsReturnCode;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|489698|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>/// &lt;summary&gt;
/// Gets a first different char occurence index
/// &lt;/summary&gt;
/// &lt;param name="a"&gt;First string&lt;/param&gt;
/// &lt;param name="b"&gt;Second string&lt;/param&gt;
/// &lt;param name="handleLengthDifference"&gt;
/// If true will return index of first occurence even strings are of different length
/// and same-length parts are equals otherwise -1
/// &lt;/param&gt;
/// &lt;returns&gt;
/// Returns first difference index or -1 if no difference is found
/// &lt;/returns&gt;
public int GetFirstBreakIndex(string a, string b, bool handleLengthDifference)
{
 int equalsReturnCode = -1;
 if (String.IsNullOrEmpty(a) || String.IsNullOrEmpty(b))
 {
 return handleLengthDifference ? 0 : equalsReturnCode;
 }

 string longest = b.Length &gt; a.Length ? b : a;
 string shorten = b.Length &gt; a.Length ? a : b; 
 for (int i = 0; i &lt; shorten.Length; i++)
 {
 if (shorten[i] != longest[i])
 {
 return i;
 }
 }

 // Handles cases when length is different (a="1234", b="123")
 // index=3 would be returned for this case
 // If you do not need such behaviour - just remove this
 if (handleLengthDifference &amp;&amp; a.Length != b.Length)
 {
 return shorten.Length;
 }

 return equalsReturnCode;
}
</code></pre>

blocks|key|705412|text|据我所知没有，但这是相当微不足道的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|705413|public+static+int+FirstUnmatchedIndex(this+string+x,+string+y)
{
++if(x+==+null+%7C%7C+y+==+null)
++++throw+new+ArgumentNullException();
++int+count+=+x.Length;
++if(count+>+y.Length)
++++return+FirstUnmatchedIndex(y,+x);
++if(ReferenceEquals(x,+y))
++++return+-1;
++for(idx+=+0;+idx+!=+count;+%2B%2Bidx)
++++if(x[idx]+!=+y[idx])
++++++return+idx;
++return+count+==+y.Length?+-1+:+count;
}|code-block|syntax|javascript|705414|这是一个简单的序数比较。顺序的不区分大小写的比较很容易更改，但区域性基础的定义很棘手；“Wei«bier”与第二个字符串中最后一个S上的"WEISSBIERS“不匹配，但这算不算位置8或位置9？|705415|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|$]]

Not that I know of, but it's pretty trivial:

<pre><code>public static int FirstUnmatchedIndex(this string x, string y)
{
 if(x == null || y == null)
 throw new ArgumentNullException();
 int count = x.Length;
 if(count &gt; y.Length)
 return FirstUnmatchedIndex(y, x);
 if(ReferenceEquals(x, y))
 return -1;
 for(idx = 0; idx != count; ++idx)
 if(x[idx] != y[idx])
 return idx;
 return count == y.Length? -1 : count;
}
</code></pre>

This is a simple ordinal comparison. Ordinal case-insensitive comparison is an easy change, but culture-base is tricky to define; "Weißbier" mismatches "WEISSBIERS" on the final S in the second string, but does that count as position 8 or position 9?

blocks|key|489796|text|可以像这样编写字符串扩展|type|unstyled|depth|inlineStyleRanges|entityRanges|data|489797|public+static+class+MyExtensions
{
++++public+static+IList<char>+Mismatch(this+string+str1,+string+str2)
++++{
++++++++var+char1+=+str1.ToCharArray();
++++++++var+char2+=+str2.ToCharArray();
++++++++IList<Char>+Resultchar=+new+List<char>();
++++++++for+(int+i+=+0;+i+<+char2.Length;i%2B%2B+)
++++++++{
++++++++++++if+(i+>=+char1.Length+%7C%7C+char1[i]+!=+char2[i])
++++++++++++++++Resultchar.Add(char2[i]);
++++++++}
++++++++return+Resultchar;
++++}
}|code-block|syntax|javascript|489798|像这样使用它|489799|var+r+=+"1234567890".Mismatch("1234567880");|489800|它不是用于查找不匹配的优化算法。|489801|如果您只对查找第一个不匹配感兴趣，|489802|public+static+Char+FirstMismatch(this+string+str1,+string+str2)
++++++++{
++++++++++++var+char1+=+str1.ToCharArray();
++++++++++++var+char2+=+str2.ToCharArray();+++++++++++++
++++++++++++for+(int+i+=+0;+i+<+char2.Length;i%2B%2B+)
++++++++++++{
++++++++++++++++if+(i+>=+char1.Length+%7C%7C+char1[i]+!=+char2[i])
++++++++++++++++++++return+char2[i];
++++++++++++}
++++++++++++return+''c;
++++++++}|489803|entityMap^0|0|0|0|0|0|0|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]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

It possible to write string extension like 

<pre><code>public static class MyExtensions
{
 public static IList&lt;char&gt; Mismatch(this string str1, string str2)
 {
 var char1 = str1.ToCharArray();
 var char2 = str2.ToCharArray();
 IList&lt;Char&gt; Resultchar= new List&lt;char&gt;();
 for (int i = 0; i &lt; char2.Length;i++ )
 {
 if (i &gt;= char1.Length || char1[i] != char2[i])
 Resultchar.Add(char2[i]);
 }
 return Resultchar;
 }
}
</code></pre>

Use it like 

<pre><code>var r = "1234567890".Mismatch("1234567880");
</code></pre>

It is not an optimized algorithm for finding the mismatch.

If you are interested only to find the the first mismatch,

<pre><code>public static Char FirstMismatch(this string str1, string str2)
 {
 var char1 = str1.ToCharArray();
 var char2 = str2.ToCharArray(); 
 for (int i = 0; i &lt; char2.Length;i++ )
 {
 if (i &gt;= char1.Length || char1[i] != char2[i])
 return char2[i];
 }
 return ''c;
 }
</code></pre>

blocks|key|489838|text|本主题是Comparing+strings+and+get+the+first+place+where+they+vary+from+eachother的副本，其中包含使用Linq的更好的单行解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|489839|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/4585939/comparing-strings-and-get-the-first-place-where-they-vary-from-eachother^0|4|20|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

This topic is a duplicate of <a href="https://stackoverflow.com/questions/4585939/comparing-strings-and-get-the-first-place-where-they-vary-from-eachother">Comparing strings and get the first place where they vary from eachother</a>, which contains a better, one line, solution using Linq

Is there an exsting string comparison method that will return a value based on the first occurance of a non matching character between two strings?

i.e. 

<pre><code>string A = "1234567890"

string B = "1234567880"
</code></pre>

I would like to get a value back that would allow me to see that the first occurance of a matching break is A[8]

c# string comparison method returning index of first non match

是否有一个现有的字符串比较方法可以根据两个字符串之间首次出现不匹配的字符来返回值？即string A = "1234567890"string B = "1234567880"我想要得到一个返回值，它允许我看到匹配中断的第一个出现是A8

问c#字符串比较方法返回第一个不匹配的索引
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