blocks|key|1840741|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1840742|Random+r+=+new+Random();
int+lvfirst+=+r.Next(485924948);
int+lvsecond+=+r.Next(39);
long+lvdone+=+Int64.Parse(lvfirst.ToString()+%2B+lvsecond.ToString());
Globals.GlobalInt+=+lvdone;|code-block|syntax|javascript|1840743|基本上，我们将所有ints转换为一个字符串，该字符串将第二个数字附加到第一个数字的末尾，然后将其转换回int。|offset|length|style|CODE|1840744|希望这能有所帮助！|1840745|entityMap^0|0|0|9|4|1F|3|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

You could do something like this:

<pre><code>Random r = new Random();
int lvfirst = r.Next(485924948);
int lvsecond = r.Next(39);
long lvdone = Int64.Parse(lvfirst.ToString() + lvsecond.ToString());
Globals.GlobalInt = lvdone;
</code></pre>

Basically we are converting all of the <code>ints</code> to a string which will append the second number to the end of the first number, then we convert it back to an <code>int</code>.

Hope this Helps!

blocks|key|3301311|text|是否要添加或连接int？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3301312|如果你想要连接，你可以尝试：|3301313|long+lvdone+=+long.Parse(lvfirst.toString()+%2B+lvsecond);|code-block|syntax|javascript|3301314|如果您想添加它们并避免溢出：|3301315|long+lvdone+=+((long)+lvfist)+%2B+lvsecond|3301316|如果数字必须任意长，请尝试使用Decimal而不是long|3301317|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Do you want to add or concatenate your ints ?

If you want to concatenate, you can try:

<pre><code>long lvdone = long.Parse(lvfirst.toString() + lvsecond);
</code></pre>

If you want to add them and avoid overflows: 

<pre><code>long lvdone = ((long) lvfist) + lvsecond
</code></pre>

If you number must be arbitrarily long, try using Decimal instead of long

blocks|key|1576255|text|这段代码有两个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1576256|1576257|将Next的两个输出“粘合在一起”不会给出目标范围内的随机值。|ordered-list-item|offset|length|style|CODE|1576258|不适用于返回结果的strings.|1576259|Instead，而是将其存储在全局变量中。|1576260|1576261|如果需要指定范围内的值，请使用|1576262|var+result+=+(long)+(r.NextDouble()+*+48592494839);|code-block|syntax|javascript|1576263|这仍然不适用于任何目标范围，但它将轻松地满足您的特定要求。|1576264|entityMap^0|0|0|1|4|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|11|8|@$F|12|G|13|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|14|8|@]|9|@]|A|$]]|$1|L|3|M|5|E|7|15|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|16|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|17|8|@]|9|@]|A|$]]|$1|Q|3|R|5|S|7|18|8|@]|9|@]|A|$T|U]]|$1|V|3|W|5|6|7|19|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|Y|$]]

There are two problems with this code:

<ol>
<li>"Sticking together" two outputs of <code>Next</code> will not give you a random value in the target range. Math does not work with strings.</li>
<li>Instead of returning a result it stores it in a global variable.</li>
</ol>

If you want a value in the range you specify, use

<pre><code>var result = (long) (r.NextDouble() * 48592494839);
</code></pre>

This will still not work for any target range, but it will comfortably meet your specific requirements.

blocks|key|3301461|text|忽略随机的东西。这是连接到数字的方式...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3301462|int+a+=+124;
int+b+=+101040507;

int+digits+=+32;
mask+=+0x80000000;
while(a+&+mask+==+0)
{
++++digits--;
++++mask=+mask+>>+1;
}


long+result+=+b;
result+=+result+<<+digits;
result+=+result+%7C+b;|code-block|syntax|javascript|3301463|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|$]]

Ignoring the random thing...
this is the way to concat to numbers...

<pre><code>int a = 124;
int b = 101040507;

int digits = 32;
mask = 0x80000000;
while(a &amp; mask == 0)
{
 digits--;
 mask= mask &gt;&gt; 1;
}


long result = b;
result = result &lt;&lt; digits;
result = result | b;
</code></pre>

It might be quite funny but I really didn't know how to search to find the answer for this one.

I always use something like this when I want to join strings "string = "something" + "somethingelse" "

but how to do it with INT? :)

<pre><code>Random r = new Random();
int lvfirst = r.Next(485924948);
int lvsecond = r.Next(39);
int lvdone = lvfirst + lvsecond;
Globals.GlobalInt = lvdone;
</code></pre>

I tried doing one long int but it doesn't seem to work it says something about Long so if you can help me how to join this 2 random numbers into 1? I need 1 random number max "48592494839"

Thanks a lot!

C# Join 2 Numbers (INT) into 1 longer INT

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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这可能很有趣，但我真的不知道如何搜索来找到这个问题的答案。当我想要连接字符串"string = " something“+ "somethingelse”“时，我总是这样使用。”但是如何使用INT呢？:)Random r = new Random();int lvfirst = r.Next(485924948);in...

问C#将2个数字(整数)合并为1个更长的整数
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