blocks|key|1046783|text|bool+IsInteger(double+num)+{
++++if+(ceil(num)+==+num+&&+floor(num)+==+num)
++++++++return+true;
++++else
++++++++return+false;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1046784|索尔沃的问题。|unstyled|1046785|编辑:+Mark+Rushakoff的Pwned。|1046786|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>bool IsInteger(double num) {
 if (ceil(num) == num &amp;&amp; floor(num) == num)
 return true;
 else
 return false;
}
</code></pre>

Problemo solvo.

Edit: Pwned by Mark Rushakoff.

blocks|key|1261750|text|有许多方法可以做到这一点。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1261751|double+d+=+5.0;
bool+isInt+=+d+==+(int)d;|code-block|syntax|javascript|1261752|你也可以使用模数。|1261753|double+d+=+5.0;
bool+isInt+=+d+%25+1+==+0;|1261754|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|$]]

There are any number of ways to do this. For example:

<pre><code>double d = 5.0;
bool isInt = d == (int)d;
</code></pre>

You can also use modulo.

<pre><code>double d = 5.0;
bool isInt = d % 1 == 0;
</code></pre>

blocks|key|1261870|text|虽然所提出的解决方案似乎对简单的示例有效，但一般来说，这样做不是一个好主意。数字可能不完全是整数，但当您尝试对其进行格式化时，它与整数非常接近，因此得到的结果是1.000000。如果您进行的计算在理论上应该正好是1，但实际上由于舍入误差而给出的数字非常接近但并不完全等于1，则可能会发生这种情况。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1261871|相反，首先格式化它，如果字符串以句点结尾，后面跟着零，那么就去掉它们。也有一些格式可以自动使用去掉尾随零。这对你的目的来说可能已经足够了。|1261872|double+d+=+1.0002;
Console.WriteLine(d.ToString("0.##"));
d+=+1.02;
Console.WriteLine(d.ToString("0.##"));|code-block|syntax|javascript|1261873|输出：|1261874|1
1.02|1261875|entityMap^0|28|8|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

Whilst the solutions proposed appear to work for simple examples, doing this in general is a bad idea. A number might not be exactly an integer but when you try to format it, it's close enough to an integer that you get <code>1.000000</code>. This can happen if you do a calculation that in theory should give exactly 1, but in practice gives a number very close to but not exactly equal to one due to rounding errors.

Instead, format it first and if your string ends in a period followed by zeros then strip them. There are also some formats that you can use that strip trailing zeros automatically. This might be good enough for your purpose.

<pre><code>double d = 1.0002;
Console.WriteLine(d.ToString("0.##"));
d = 1.02;
Console.WriteLine(d.ToString("0.##"));
</code></pre>

Output:

<pre><code>1
1.02
</code></pre>

blocks|key|1047031|text|可以对double类型使用字符串格式。下面是一个示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1047032|double+val+=+58.6547;
String.Format("{0:0.##}",+val);++++++
//Output:+"58.65"

double+val+=+58.6;
String.Format("{0:0.##}",+val);++++++
//Output:+"58.6"

double+val+=+58.0;
String.Format("{0:0.##}",+val);++++++
//Output:+"58"|code-block|syntax|javascript|1047033|如果这不管用，请告诉我。|1047034|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|$]]

You can use String formatting for the double type. Here is an example:

<pre><code>double val = 58.6547;
String.Format("{0:0.##}", val); 
//Output: "58.65"

double val = 58.6;
String.Format("{0:0.##}", val); 
//Output: "58.6"

double val = 58.0;
String.Format("{0:0.##}", val); 
//Output: "58"
</code></pre>

Let me know if this doesn't help.

blocks|key|1261985|text|Mark+Rushakoff的答案可能更简单，但以下方法也适用，并且可能更有效，因为没有隐式除法运算：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1261986|+++++bool+isInteger+=+(double)((int)f)+==+f+;|code-block|syntax|javascript|1261987|和|1261988|+++++bool+isInteger+=+(decimal)((int)d)+==+d+;|1261989|如果您想为两种类型都使用一个表达式，也许|1261990|+++++bool+isInteger+=+(double)((int)val)+==+(double)val+;|1261991|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Mark Rushakoff's answer may be simpler, but the following also work and may be more efficient since there is no implicit division operation:

<pre><code> bool isInteger = (double)((int)f) == f ;
</code></pre>

and

<pre><code> bool isInteger = (decimal)((int)d) == d ;
</code></pre>

If you want a single expression for both types, perhaps

<pre><code> bool isInteger = (double)((int)val) == (double)val ;
</code></pre>

blocks|key|1262057|text|我遇到过类似的情况，但值是一个字符串。用户输入一个应该是美元金额的值，所以我想验证它是数字的，并且最多有两个小数位。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1262058|下面是我的代码，如果字符串"s“表示一个最多有两位小数的数字，则返回true，否则返回false。它避免了由于浮点值的不精确而导致的任何问题。|1262059|try
{
++++//+must+be+numeric+value
++++double+d+=+double.Parse(s);
++++//+max+of+two+decimal+places
++++if+(s.IndexOf(".")+>=+0)
++++{
++++++++if+(s.Length+>+s.IndexOf(".")+%2B+3)
++++++++++++return+false;
++++}
++++return+true;
catch
{
++++return+false;
}|code-block|syntax|javascript|1262060|我在http://progblog10.blogspot.com/2011/04/determining-whether-numeric-value-has.html上对此进行了更详细的讨论。|offset|length|1262061|entityMap|0|LINK|mutability|MUTABLE|url|http://progblog10.blogspot.com/2011/04/determining-whether-numeric-value-has.html^0|0|0|0|2|29|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]]|$1|I|3|J|5|6|7|X|8|@]|9|@$K|Y|L|Z|1|10]]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

I faced a similar situation, but where the value is a string. The user types in a value that's supposed to be a dollar amount, so I want to validate that it's numeric and has at most two decimal places.

Here's my code to return true if the string "s" represents a numeric with at most two decimal places, and false otherwise. It avoids any problems that would result from the imprecision of floating-point values.

<pre><code>try
{
 // must be numeric value
 double d = double.Parse(s);
 // max of two decimal places
 if (s.IndexOf(".") &gt;= 0)
 {
 if (s.Length &gt; s.IndexOf(".") + 3)
 return false;
 }
 return true;
catch
{
 return false;
}
</code></pre>

I discuss this in more detail at <a href="http://progblog10.blogspot.com/2011/04/determining-whether-numeric-value-has.html" rel="nofollow">http://progblog10.blogspot.com/2011/04/determining-whether-numeric-value-has.html</a>.

blocks|key|1047202|text|如果Int32的上下限很重要：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1047203|public+bool+IsInt32(double+value)
{
++++return++value+>=+int.MinValue+&&+value+<=+int.MaxValue+&&+value+==+(int)value;
}|code-block|syntax|javascript|1047204|entityMap^0|2|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If upper and lower bound of <code>Int32</code> matters:

<pre><code>public bool IsInt32(double value)
{
 return value &gt;= int.MinValue &amp;&amp; value &lt;= int.MaxValue &amp;&amp; value == (int)value;
}
</code></pre>

blocks|key|1047271|text|也许不是最优雅的解决方案，但如果你不太挑剔的话，它是有效的！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1047272|bool+IsInteger(double+num)+{
++++return+!num.ToString("0.################").Contains(".");
}|code-block|syntax|javascript|1047273|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|$]]

Perhaps not the most elegant solution but it works if you are not too picky!

<pre><code>bool IsInteger(double num) {
 return !num.ToString("0.################").Contains(".");
}
</code></pre>

blocks|key|1047290|text|static+bool+IsWholeNumber(double+x)+
{
++++return+Math.Abs(x+%25+1)+<+double.Epsilon;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1047291|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>static bool IsWholeNumber(double x) 
{
 return Math.Abs(x % 1) &lt; double.Epsilon;
}
</code></pre>

blocks|key|1047296|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1047297|number+==+Convert.ToInt16(number);|code-block|syntax|javascript|1047298|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|$]]

Try this:

<pre><code>number == Convert.ToInt16(number);
</code></pre>

blocks|key|1047343|text|使用int.TryParse将产生以下结果：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1047344|++++++++var+shouldBeInt+=+3;

++++++++var+shouldntBeInt+=+3.1415;

++++++++var+iDontWantThisToBeInt+=+3.000f;

++++++++Console.WriteLine(int.TryParse(shouldBeInt.ToString(),+out+int+parser));+//+true

++++++++Console.WriteLine(int.TryParse(shouldntBeInt.ToString(),+out+parser));+//+false

++++++++Console.WriteLine(int.TryParse(iDontWantThisToBeInt.ToString(),+out+parser));+//+true,+even+if+I+don't+want+this+to+be+int

++++++++Console.WriteLine(int.TryParse("3.1415",+out++parser));+//+false

++++++++Console.WriteLine(int.TryParse("3.0000",+out+parser));+//+false

++++++++Console.WriteLine(int.TryParse("3",+out+parser));+//+true

++++++++Console.ReadKey();|code-block|syntax|javascript|1047345|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|$]]

Using int.TryParse will yield these results:

<pre><code> var shouldBeInt = 3;

 var shouldntBeInt = 3.1415;

 var iDontWantThisToBeInt = 3.000f;

 Console.WriteLine(int.TryParse(shouldBeInt.ToString(), out int parser)); // true

 Console.WriteLine(int.TryParse(shouldntBeInt.ToString(), out parser)); // false

 Console.WriteLine(int.TryParse(iDontWantThisToBeInt.ToString(), out parser)); // true, even if I don't want this to be int

 Console.WriteLine(int.TryParse("3.1415", out parser)); // false

 Console.WriteLine(int.TryParse("3.0000", out parser)); // false

 Console.WriteLine(int.TryParse("3", out parser)); // true

 Console.ReadKey();
</code></pre>

blocks|key|1047356|text|++++public+static+bool+isInteger(decimal+n)
++++{
++++++++return+n+-+(Int64)n+==+0;
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1047357|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> public static bool isInteger(decimal n)
 {
 return n - (Int64)n == 0;
 }
</code></pre>

blocks|key|1047394|text|您可以简单地将double与相同值的int类型转换进行比较。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1047395|double+d+=+5.0m;

if+(d+==+(int)d)
{
....+
}|code-block|syntax|javascript|1047396|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|$]]

You can simply compare the double against the int cast of the same value.
<pre><code>double d = 5.0m;

if (d == (int)d)
{
.... 
}
</code></pre>

How do I tell if a decimal or double value is an integer?

For example:

<pre><code>decimal d = 5.0; // Would be true
decimal f = 5.5; // Would be false
</code></pre>

or 

<pre><code>double d = 5.0; // Would be true
double f = 5.5; // Would be false
</code></pre>

The reason I would like to know this is so that I can determine programmatically if I want to output the value using <code>.ToString("N0")</code> or <code>.ToString("N2")</code>. If there is no decimal point value, then I don't want to show that.

How to determine if a decimal/double is an integer?

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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如何判断十进制值或双精度值是否为整数？例如：decimal d = 5.0; // Would be truedecimal f = 5.5; // Would be false或double d = 5.0; // Would be truedouble f = 5.5; // Would be false我之所以想...

问如何判断小数/双精度数是否为整数？
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问如何判断小数/双精度数是否为整数？EN