convert double to int

Question

What is the best way to convert a double to an int? Should a cast be used?

@Armen - touche` :) Doesn't really matter anyway - Skeet is here, nothing else matters. — RPM1984, Nov 15 '10 at 06:42

score 274 · Accepted Answer · answered Nov 15 '10 at 06:30

274

You can use a cast if you want the default truncate-towards-zero behaviour. Alternatively, you might want to use Math.Ceiling, Math.Round, Math.Floor etc - although you'll still need a cast afterwards.

Don't forget that the range of int is much smaller than the range of double. A cast from double to int won't throw an exception if the value is outside the range of int in an unchecked context, whereas a call to Convert.ToInt32(double) will. The result of the cast (in an unchecked context) is explicitly undefined if the value is outside the range.

answered Nov 15 '10 at 06:30

Jon Skeet

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I am wondering if in 64 bits machine, Convert.ToInt32 is not OK any more? – user496949 Nov 15 '10 at 08:23
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@user: The size of an `int` is always 32 bits, regardless of whether you're using a 32 or 64 bit machine. – Joren Nov 15 '10 at 10:15
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And doubles can be much huger than even a 64 bit int. – Adrian Ratnapala May 01 '13 at 08:07
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For C# noobs like me: both Math and Convert are part of System. So the complete answer looks like this : `intVal = System.Convert.ToInt32(System.Math.Floor(dblVal));` – kris May 01 '16 at 08:07
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@user1290746: That would work, but you'd usually just have `using System;` at the top of the file, at which point it could just be `intVal = Convert.ToInt32(Math.Floor(dblVal));` – Jon Skeet May 01 '16 at 08:08
Obviously that is obvious to anyone familiar with C#. And someone searching for an answer about type conversion is possibly/likely completely unfamiliar with the language - that was the case for me anyway - so thanks for your extra info. – kris May 01 '16 at 08:14
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@user1290746: But in that case, I don't think the right solution is to advise people to use the fully-qualified names - the better solution is to learn about `using` directives. – Jon Skeet May 01 '16 at 08:15
Interesting that in VB.Net Double to Integer cast like `CInt(Double)` compiles to the default `Math.Round(Double)` call and `conv.ovf.i4`, but in C# `(int)double` compiles to `conv.i4`. In VB.Net `CInt(1.5)` is 2, but in C# `(int)1.5` is 1. – Slai Nov 21 '16 at 13:55

score 47 · Answer 2 · answered Nov 15 '10 at 06:29

47

Yeah, why not?

double someDouble = 12323.2;
int someInt = (int)someDouble;

Using the Convert class works well too.

int someOtherInt = Convert.ToInt32(someDouble);

answered Nov 15 '10 at 06:29

Jeff Mercado

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score 44 · Answer 3 · answered Nov 15 '10 at 06:30

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if you use cast, that is, (int)SomeDouble you will truncate the fractional part. That is, if SomeDouble were 4.9999 the result would be 4, not 5. Converting to int doesn't round the number. If you want rounding use Math.Round

answered Nov 15 '10 at 06:30

Armen Tsirunyan

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score 8 · Answer 4 · edited Mar 10 '15 at 18:11

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The best way is to simply use Convert.ToInt32. It is fast and also rounds correctly.

Why make it more complicated?

edited Mar 10 '15 at 18:11

Jeff B

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answered May 30 '13 at 01:59

majed

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You are right. The only important point to pay additional attention is what Jon Skeet has written: "Don't forget that the range of int is much smaller than the range of double" – Mickey Pearson Jul 24 '23 at 09:10

score 6 · Answer 5 · answered Nov 15 '10 at 06:30

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Convert.ToInt32 is the best way to convert

answered Nov 15 '10 at 06:30

anishMarokey

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This a typically unsafe function because it accepts just anything. – Anton Shepelev Jul 05 '16 at 14:48

score 3 · Answer 6 · answered Nov 15 '10 at 06:37

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My ways are :

 - Convert.ToInt32(double_value)
 - (int)double_value
 - Int32.Parse(double_value.ToString());

answered Nov 15 '10 at 06:37

kst

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Don't you think the third way is unnecessarily complex and slow? – Armen Tsirunyan Nov 15 '10 at 06:38
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The third will not work with doubles with fractional parts. e.g., `double_value = 0.1` – Jeff Mercado Nov 15 '10 at 06:46
I think u can use Math.Floor. – lindexi Apr 24 '17 at 02:02

score 2 · Answer 7 · edited Aug 20 '21 at 09:07

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None of the code samples exhibit the normal behaviour a school kid would expect (that halves are rounded up). What most people would expect is is this:

double d = 1.5;
int i = (int)Math.Round(d, MidpointRounding.AwayFromZero);

edited Aug 20 '21 at 09:07

Dharman

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answered Aug 20 '21 at 09:01

RAM

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score 1 · Answer 8 · edited Nov 15 '10 at 07:38

Here is a complete example

class Example 
{    
  public static void Main() 
  {    
    double x, y; 
    int i; 

    x = 10.0; 
    y = 3.0; 

    // cast double to int, fractional component lost (Line to be replaced) 
    i = (int) (x / y); 
    Console.WriteLine("Integer outcome of x / y: " + i); 
  }    
}

If you want to round the number to the closer integer do the following:

i = (int) Math.Round(x / y); // Line replaced

score 1 · Answer 9 · edited Mar 05 '22 at 18:10

There is no better. It really depends on what you want, performance, accuracy, etc.

But see differences https://dotnetfiddle.net/kwqF2M

double testeDouble_min = 12.3456;
double testeDouble_max = 12.8456;
double emptyDouble;

int i01_min = Convert.ToInt32(testeDouble_min); //12
int i01_max = Convert.ToInt32(testeDouble_max); //13
//var erro = Convert.ToInt32(emptyDouble); //error  

int i02_min = (int)testeDouble_min; //12
int i02_max = (int)testeDouble_max; //12
//var erro = (int)emptyDouble; //error  

var styles = System.Globalization.NumberStyles.Integer | System.Globalization.NumberStyles.AllowDecimalPoint;
var provider = new System.Globalization.CultureInfo("en-US");
bool success = int.TryParse(testeDouble_min.ToString(), styles, provider, out int i03_min); //0

_ = int.TryParse(testeDouble_max.ToString(),  out int i03_max); //0 because it has decimal place, if it didn't have ok

int i04_min = (int)Math.Round(testeDouble_min, MidpointRounding.AwayFromZero); //12
int i04_max = (int)Math.Round(testeDouble_max, MidpointRounding.AwayFromZero); //13

About IL there are no big differences between (int) or Convert.ToInt32

var dd = 12.3;
int a = (int)dd;

result:

    IL_0000: nop
    IL_0001: ldc.r8 12.3
    IL_000a: stloc.0
    IL_000b: ldloc.0
    IL_000c: conv.i4
    IL_000d: stloc.1
    IL_000e: ret

(int) in theory less instructions tend to be faster, but I doubt anyone can measure that.

And

var dd = 12.3;
int b = Convert.ToInt32(dd);

result:

    IL_0000: nop
    IL_0001: ldc.r8 12.3
    IL_000a: stloc.0
    IL_000b: ldloc.0
    IL_000c: call int32 [System.Private.CoreLib]System.Convert::ToInt32(float64)
    IL_0011: stloc.1
    IL_0012: ret

score 0 · Answer 10 · answered Apr 19 '17 at 23:21

0

int myInt = (int)Math.Ceiling(myDouble);

answered Apr 19 '17 at 23:21

mathewsun

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zangetsu · Answer 11 · 2023-01-30T17:34:19.010

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Use the inbuilt Convert class. It offers all type of conversion methods whether from int to double, double to float or byte etc.

int i= Convert.ToInt32(500*3.142456778)

edited Jan 30 '23 at 17:34

answered Jan 30 '23 at 17:33

zangetsu

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score -1 · Answer 12 · answered Jul 14 '14 at 18:59

-1

label8.Text = "" + years.ToString("00") + " years";

when you want to send it to a label, or something, and you don't want any fractional component, this is the best way

label8.Text = "" + years.ToString("00.00") + " years";

if you want with only 2, and it's always like that

answered Jul 14 '14 at 18:59

Ruth

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This is most certainly **not the best way** and you should probably retract your answer. By using `ToString` you're newing up a lot of other costly resources behind the scenes, like a string formatter and parser which are meant for special purposes. Learn from the answers given here. The closest to best way is to cast to `int` and truncate the fractional part or use something like `Math.Round` to get to the closest `int` you're looking for. – Paul-Sebastian Manole Jul 10 '21 at 22:05

score -1 · Answer 13 · edited Aug 18 '12 at 23:39

-1

I think the best way is Convert.ToInt32.

edited Aug 18 '12 at 23:39

Christoffer Lette

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answered Nov 15 '10 at 07:24

Singleton

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convert double to int

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