entityMap|blocks|key|1vjpn|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|lhak|string+s+=+inputValue.ToString("0.00",+CultureInfo.InvariantCulture);
string[]+parts+=+s.Split('.');+
int+i1+=+int.Parse(parts[0]);
int+i2+=+int.Parse(parts[1]);|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

This is how you could do it:

<pre><code>string s = inputValue.ToString("0.00", CultureInfo.InvariantCulture);
string[] parts = s.Split('.'); 
int i1 = int.Parse(parts[0]);
int i2 = int.Parse(parts[1]);
</code></pre>

entityMap|blocks|key|2oi91|text|操纵字符串可能会很慢。尝试使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|8gu92|double+number;

long+intPart+=+(long)+number;
double+fractionalPart+=+number+-+intPart;|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

Manipulating strings can be slow. Try using the following:

<pre><code>double number;

long intPart = (long) number;
double fractionalPart = number - intPart;
</code></pre>

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://en.wikipedia.org/wiki/Modulo_operation|blocks|key|1rhuh|text|你想用什么编程语言来做这件事？大多数语言都应该有一个模运算符..。C%2B%2B示例：|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|syntax|javascript|2br8j|double+num+=+10.5;
int+remainder+=+num+%25+1|code-block^0|Q|5|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|P|G|@]|H|@$I|Q|J|R|A|S]]|6|$K|L]]|$A|M|C|N|2|O|F|T|G|@]|H|@]|6|$K|L]]]]

What programming language you want to use to do this? Most of the language should have a <a href="http://en.wikipedia.org/wiki/Modulo_operation" rel="noreferrer">Modulo operator</a>. C++ example:

<pre><code>double num = 10.5;
int remainder = num % 1
</code></pre>

entityMap|blocks|key|7s3i7|text|"10.50".Split('.').Select(int.Parse);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|js^0^^$0|$]|1|@$2|3|4|5|6|7|8|E|9|@]|A|@]|B|$C|D]]]]

<pre><code>"10.50".Split('.').Select(int.Parse);
</code></pre>

entityMap|blocks|key|e2f49|text|///+
///+Get+the+integral+and+floating+point+portions+of+a+Double
///+as+separate+integer+values,+where+the+floating+point+value+is+
///+raised+to+the+specified+power+of+ten,+given+by+'places'.
///+
public+static+void+Split(Double+value,+Int32+places,+out+Int32+left,+out+Int32+right)
{
++++left+=+(Int32)Math.Truncate(value);
++++right+=+(Int32)((value+-+left)+*+Math.Pow(10,+places));
}

public+static+void+Split(Double+value,+out+Int32+left,+out+Int32+right)
{
++++Split(value,+1,+out+left,+out+right);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4n8jr|用法：|unstyled|7i748|Int32+left,+right;

Split(10.50,+out+left,+out+right);
//+left+==+10
//+right+==+5

Split(10.50,+2,+out+left,+out+right);
//+left+==+10
//+right+==+50

Split(10.50,+5,+out+left,+out+right);
//+left+==+10
//+right+==+50000^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|J|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|K|9|@]|A|@]|B|$]]|$2|H|4|I|6|7|8|L|9|@]|A|@]|B|$C|D]]]]

<pre><code>/// &lt;summary&gt;
/// Get the integral and floating point portions of a Double
/// as separate integer values, where the floating point value is 
/// raised to the specified power of ten, given by 'places'.
/// &lt;/summary&gt;
public static void Split(Double value, Int32 places, out Int32 left, out Int32 right)
{
 left = (Int32)Math.Truncate(value);
 right = (Int32)((value - left) * Math.Pow(10, places));
}

public static void Split(Double value, out Int32 left, out Int32 right)
{
 Split(value, 1, out left, out right);
}
</code></pre>

Usage:

<pre><code>Int32 left, right;

Split(10.50, out left, out right);
// left == 10
// right == 5

Split(10.50, 2, out left, out right);
// left == 10
// right == 50

Split(10.50, 5, out left, out right);
// left == 10
// right == 50000
</code></pre>

entityMap|blocks|key|7fir3|text|怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4tf5j|var+n+=+1004.522
var+a+=+Math.Floor(n);
var+b+=+n+-+a;|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

how about?

<pre><code>var n = 1004.522
var a = Math.Floor(n);
var b = n - a;
</code></pre>

entityMap|blocks|key|d596i|text|另一个不涉及字符串操作的变体：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|f8mks|static+void+Main(string[]+args)
{
++++decimal+number+=+10123.51m;
++++int+whole+=+(int)number;
++++decimal+precision+=+(number+-+whole)+*+100;

++++Console.WriteLine(number);
++++Console.WriteLine(whole);
++++Console.WriteLine("{0}+and+{1}",whole,(int)+precision);
++++Console.Read();
}|code-block|syntax|javascript|a5nke|确保它们是小数，否则你会得到通常奇怪的浮点数/双精度数行为。^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|J|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|K|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|L|9|@]|A|@]|B|$]]]]

Another variation that doesn't involve string manipulation:

<pre><code>static void Main(string[] args)
{
 decimal number = 10123.51m;
 int whole = (int)number;
 decimal precision = (number - whole) * 100;

 Console.WriteLine(number);
 Console.WriteLine(whole);
 Console.WriteLine("{0} and {1}",whole,(int) precision);
 Console.Read();
}
</code></pre>

Make sure they're decimals or you get the usual strange float/double behaviour.

entityMap|blocks|key|a11eu|text|你可以用字符串拆分，然后转换成整型...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6mhbh|string+s+=+input.ToString();+
string[]+parts+=+s.Split('.');|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

you can split with string and then convert into int ...

<pre><code>string s = input.ToString(); 
string[] parts = s.Split('.');
</code></pre>

entityMap|blocks|key|dp2fs|text|这个函数需要十进制的时间，然后转换回基数60。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1tv1i|public+string+Time_In_Absolute(double+time)
++++{
++++++++time+=+Math.Round(time,+2);
++++++++string[]+timeparts+=+time.ToString().Split('.');++++++++++++++++++++++++
++++++++timeparts[1]+=+"."+%2B+timeparts[1];
++++++++double+Minutes+=+double.Parse(timeparts[1]);++++++++++++
++++++++Minutes+=+Math.Round(Minutes,+2);
++++++++Minutes+=+Minutes+*+(double)60;
++++++++return+string.Format("{0:00}:{1:00}",timeparts[0],Minutes);
++++++++//return+Hours.ToString()+%2B+":"+%2B+Math.Round(Minutes,0).ToString();+
++++}|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

This function will take time in decimal and converts back into base 60 .

<pre><code> public string Time_In_Absolute(double time)
 {
 time = Math.Round(time, 2);
 string[] timeparts = time.ToString().Split('.'); 
 timeparts[1] = "." + timeparts[1];
 double Minutes = double.Parse(timeparts[1]); 
 Minutes = Math.Round(Minutes, 2);
 Minutes = Minutes * (double)60;
 return string.Format("{0:00}:{1:00}",timeparts[0],Minutes);
 //return Hours.ToString() + ":" + Math.Round(Minutes,0).ToString(); 
 }
</code></pre>

entityMap|blocks|key|78aan|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|bop36|string+s+=+"10.5";
string[]+s1+=+s.Split(new+char[]+{+"."+});
string+first+=+s1[0];
string+second+=+s1[1];|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

Try:

<pre><code>string s = "10.5";
string[] s1 = s.Split(new char[] { "." });
string first = s1[0];
string second = s1[1];
</code></pre>

entityMap|blocks|key|95jbh|text|您可以在不使用字符串的情况下完成此操作。示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|uo54|foreach+(double+x+in+new+double[]{10.45,+10.50,+10.999,+-10.323,+-10.326,+10}){
++++int+i+=+(int)Math.Truncate(x);
++++int+f+=+(int)Math.Round(100*Math.Abs(x-i));
++++if+(f==100){+f=0;+i%2B=(x<0)?-1:1;+}
++++Console.WriteLine("("%2Bi%2B",+"%2Bf%2B")");
}|code-block|syntax|javascript|12unj|输出：|66og0|(10,+45)
(10,+50)
(11,+0)
(-10,+32)
(-10,+33)
(10,+0)|78mlt|对于像这样的数字不起作用-0.123不过。再说一次，我不确定它是否适合你的陈述。|offset|length|style|CODE^0|0|0|0|0|C|6^^$0|$]|1|@$2|3|4|5|6|7|8|R|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|S|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|T|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|U|9|@]|A|@]|B|$F|G]]|$2|L|4|M|6|7|8|V|9|@$N|W|O|X|P|Q]]|A|@]|B|$]]]]

You can do it without going through strings. Example:

<pre><code>foreach (double x in new double[]{10.45, 10.50, 10.999, -10.323, -10.326, 10}){
 int i = (int)Math.Truncate(x);
 int f = (int)Math.Round(100*Math.Abs(x-i));
 if (f==100){ f=0; i+=(x&lt;0)?-1:1; }
 Console.WriteLine("("+i+", "+f+")");
}
</code></pre>

Output:

<pre><code>(10, 45)
(10, 50)
(11, 0)
(-10, 32)
(-10, 33)
(10, 0)
</code></pre>

Won't work for a number like <code>-0.123</code>, though. Then again, I'm not sure how it would fit your representation.

entityMap|blocks|key|40lrf|text|实际上，我不得不在现实世界中回答这个问题，虽然@David+Samuel的回答做到了其中的一部分，但这里是我使用的结果代码。如前所述，字符串的开销太大了。我必须对视频中的像素值进行这种计算，并且在一台普通的计算机上仍然能够保持30fps。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5ph02|double+number+=+4140+/+640;+//result+is+6.46875+for+example

int+intPart+=+(int)number;+//just+convert+to+int,+loose+the+dec.
int+fractionalPart+=+(int)((position+-+intPart)+*+1000);+//rounding+was+not+needed.
//this+procedure+will+create+two+variables+used+to+extract+[iii*].[iii]*+from+iii*.iii*|code-block|syntax|javascript|2u1lo|这用于从640X480视频提要中的像素数求解x，y。^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|J|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|K|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|L|9|@]|A|@]|B|$]]]]

I actually just had to answer this in the real world and while @David Samuel's answer did part of it here is the resulting code I used. As said before Strings are way too much overhead. I had to do this calculation across pixel values in a video and was still able to maintain 30fps on a moderate computer. 

<pre><code>double number = 4140 / 640; //result is 6.46875 for example

int intPart = (int)number; //just convert to int, loose the dec.
int fractionalPart = (int)((position - intPart) * 1000); //rounding was not needed.
//this procedure will create two variables used to extract [iii*].[iii]* from iii*.iii*
</code></pre>

This was used to solve x,y from pixel count in 640 X 480 video feed.

entityMap|blocks|key|ktg1|text|使用Linq.只是澄清@Denis的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|8f56g|var+splt+=+"10.50".Split('.').Select(int.Parse);
int+i1+=+splt.ElementAt(0);
int+i2+=+splt.ElementAt(2);|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

Using Linq. Just clarification of @Denis answer.
<pre><code>var parts = &quot;10.50&quot;.Split('.').Select(int.Parse);
int i1 = parts.ElementAt(0);
int i2 = parts.ElementAt(1);
</code></pre>

entityMap|blocks|key|fleee|text|Console.Write("Enter+the+amount+of+money:+");
double+value+=+double.Parse(Console.ReadLine());

int+wholeDigits+=+(int)+value;
double+fractionalDigits+=+(value+-+wholeDigits)+*+100;
fractionalDigits+=+(int)+fractionalDigits;

Console.WriteLine(
++++"The+number+of+the+shekels+is+{0},+and+the+number+of+the+agurot+is+{1}",
++++wholeDigits,+fractionalDigits);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript^0^^$0|$]|1|@$2|3|4|5|6|7|8|E|9|@]|A|@]|B|$C|D]]]]

<pre><code>Console.Write(&quot;Enter the amount of money: &quot;);
double value = double.Parse(Console.ReadLine());

int wholeDigits = (int) value;
double fractionalDigits = (value - wholeDigits) * 100;
fractionalDigits = (int) fractionalDigits;

Console.WriteLine(
 &quot;The number of the shekels is {0}, and the number of the agurot is {1}&quot;,
 wholeDigits, fractionalDigits);
</code></pre>

I need to split an double value, into two int value, one before the decimal point and one after. The int after the decimal point should have two digits.

Example:

<pre><code> 10.50 = 10 and 50
 10.45 = 10 and 45
 10.5 = 10 and 50
</code></pre>

Split double into two int, one int before decimal point and one after

 我需要将一个双精度值拆分为两个整数值，一个在小数点之前，一个在小数点之后。小数点后的int应该有两位数。 示例： 10.50 = 10 and 50    10.45 = 10 and 45    10.5  = 10 and 50

问将double拆分为两个int，一个int在小数点之前，一个在小数点之后
EN

回答 14

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问将double拆分为两个int，一个int在小数点之前，一个在小数点之后EN

回答 14

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问将double拆分为两个int，一个int在小数点之前，一个在小数点之后
EN