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Type int? vs type int


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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








39















I've this comparison which equals false as expected



bool eq = typeof(int?).Equals(typeof(int));


now I have this code



List<object> items = new List<object>() (int?)123 ;
int result = items.OfType<int>().FirstOrDefault();


but this returns 123 - anyway that value is of type int?



How can this be?










share|improve this question






















  • int? boxed as int , and basically every Nullable type, Edit : Marc Gravell have the full answer

    – styx
    Mar 27 at 8:30












  • Related post about nullable type: stackoverflow.com/questions/4028830/…. This is called as "type lifting".

    – Tetsuya Yamamoto
    Mar 27 at 8:32











  • Before reading this topic I wouldn't even guess that even List<int?> already holds just int types. Proof

    – Sinatr
    Mar 27 at 8:46






  • 7





    @Sinatr no, that is incorrect; List<int?> holds int?. The important distinction in this example is the use of List<object>. What you're seeing in that "proof" is something very different; GetType() on any T? either returns the T, or throws a NRE. It never returns T? - better example: dotnetfiddle.net/3Gy3Fa - and as for why: because GetType() is non-virtual, it cannot be overridden, and thus calling GetType() is a boxing operation (even if used via "constrained call"). And when you box a T?, you either get a T as an object, or a null.

    – Marc Gravell
    Mar 27 at 8:49












  • @TetsuyaYamamoto That is not a relevant link. How is the boxing/unboxing behavior of Nullable<T> possible? is.

    – GSerg
    Mar 27 at 11:17

















39















I've this comparison which equals false as expected



bool eq = typeof(int?).Equals(typeof(int));


now I have this code



List<object> items = new List<object>() (int?)123 ;
int result = items.OfType<int>().FirstOrDefault();


but this returns 123 - anyway that value is of type int?



How can this be?










share|improve this question






















  • int? boxed as int , and basically every Nullable type, Edit : Marc Gravell have the full answer

    – styx
    Mar 27 at 8:30












  • Related post about nullable type: stackoverflow.com/questions/4028830/…. This is called as "type lifting".

    – Tetsuya Yamamoto
    Mar 27 at 8:32











  • Before reading this topic I wouldn't even guess that even List<int?> already holds just int types. Proof

    – Sinatr
    Mar 27 at 8:46






  • 7





    @Sinatr no, that is incorrect; List<int?> holds int?. The important distinction in this example is the use of List<object>. What you're seeing in that "proof" is something very different; GetType() on any T? either returns the T, or throws a NRE. It never returns T? - better example: dotnetfiddle.net/3Gy3Fa - and as for why: because GetType() is non-virtual, it cannot be overridden, and thus calling GetType() is a boxing operation (even if used via "constrained call"). And when you box a T?, you either get a T as an object, or a null.

    – Marc Gravell
    Mar 27 at 8:49












  • @TetsuyaYamamoto That is not a relevant link. How is the boxing/unboxing behavior of Nullable<T> possible? is.

    – GSerg
    Mar 27 at 11:17













39












39








39


3






I've this comparison which equals false as expected



bool eq = typeof(int?).Equals(typeof(int));


now I have this code



List<object> items = new List<object>() (int?)123 ;
int result = items.OfType<int>().FirstOrDefault();


but this returns 123 - anyway that value is of type int?



How can this be?










share|improve this question














I've this comparison which equals false as expected



bool eq = typeof(int?).Equals(typeof(int));


now I have this code



List<object> items = new List<object>() (int?)123 ;
int result = items.OfType<int>().FirstOrDefault();


but this returns 123 - anyway that value is of type int?



How can this be?







c# casting






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 27 at 8:27









Dr. SnailDr. Snail

768828




768828












  • int? boxed as int , and basically every Nullable type, Edit : Marc Gravell have the full answer

    – styx
    Mar 27 at 8:30












  • Related post about nullable type: stackoverflow.com/questions/4028830/…. This is called as "type lifting".

    – Tetsuya Yamamoto
    Mar 27 at 8:32











  • Before reading this topic I wouldn't even guess that even List<int?> already holds just int types. Proof

    – Sinatr
    Mar 27 at 8:46






  • 7





    @Sinatr no, that is incorrect; List<int?> holds int?. The important distinction in this example is the use of List<object>. What you're seeing in that "proof" is something very different; GetType() on any T? either returns the T, or throws a NRE. It never returns T? - better example: dotnetfiddle.net/3Gy3Fa - and as for why: because GetType() is non-virtual, it cannot be overridden, and thus calling GetType() is a boxing operation (even if used via "constrained call"). And when you box a T?, you either get a T as an object, or a null.

    – Marc Gravell
    Mar 27 at 8:49












  • @TetsuyaYamamoto That is not a relevant link. How is the boxing/unboxing behavior of Nullable<T> possible? is.

    – GSerg
    Mar 27 at 11:17

















  • int? boxed as int , and basically every Nullable type, Edit : Marc Gravell have the full answer

    – styx
    Mar 27 at 8:30












  • Related post about nullable type: stackoverflow.com/questions/4028830/…. This is called as "type lifting".

    – Tetsuya Yamamoto
    Mar 27 at 8:32











  • Before reading this topic I wouldn't even guess that even List<int?> already holds just int types. Proof

    – Sinatr
    Mar 27 at 8:46






  • 7





    @Sinatr no, that is incorrect; List<int?> holds int?. The important distinction in this example is the use of List<object>. What you're seeing in that "proof" is something very different; GetType() on any T? either returns the T, or throws a NRE. It never returns T? - better example: dotnetfiddle.net/3Gy3Fa - and as for why: because GetType() is non-virtual, it cannot be overridden, and thus calling GetType() is a boxing operation (even if used via "constrained call"). And when you box a T?, you either get a T as an object, or a null.

    – Marc Gravell
    Mar 27 at 8:49












  • @TetsuyaYamamoto That is not a relevant link. How is the boxing/unboxing behavior of Nullable<T> possible? is.

    – GSerg
    Mar 27 at 11:17
















int? boxed as int , and basically every Nullable type, Edit : Marc Gravell have the full answer

– styx
Mar 27 at 8:30






int? boxed as int , and basically every Nullable type, Edit : Marc Gravell have the full answer

– styx
Mar 27 at 8:30














Related post about nullable type: stackoverflow.com/questions/4028830/…. This is called as "type lifting".

– Tetsuya Yamamoto
Mar 27 at 8:32





Related post about nullable type: stackoverflow.com/questions/4028830/…. This is called as "type lifting".

– Tetsuya Yamamoto
Mar 27 at 8:32













Before reading this topic I wouldn't even guess that even List<int?> already holds just int types. Proof

– Sinatr
Mar 27 at 8:46





Before reading this topic I wouldn't even guess that even List<int?> already holds just int types. Proof

– Sinatr
Mar 27 at 8:46




7




7





@Sinatr no, that is incorrect; List<int?> holds int?. The important distinction in this example is the use of List<object>. What you're seeing in that "proof" is something very different; GetType() on any T? either returns the T, or throws a NRE. It never returns T? - better example: dotnetfiddle.net/3Gy3Fa - and as for why: because GetType() is non-virtual, it cannot be overridden, and thus calling GetType() is a boxing operation (even if used via "constrained call"). And when you box a T?, you either get a T as an object, or a null.

– Marc Gravell
Mar 27 at 8:49






@Sinatr no, that is incorrect; List<int?> holds int?. The important distinction in this example is the use of List<object>. What you're seeing in that "proof" is something very different; GetType() on any T? either returns the T, or throws a NRE. It never returns T? - better example: dotnetfiddle.net/3Gy3Fa - and as for why: because GetType() is non-virtual, it cannot be overridden, and thus calling GetType() is a boxing operation (even if used via "constrained call"). And when you box a T?, you either get a T as an object, or a null.

– Marc Gravell
Mar 27 at 8:49














@TetsuyaYamamoto That is not a relevant link. How is the boxing/unboxing behavior of Nullable<T> possible? is.

– GSerg
Mar 27 at 11:17





@TetsuyaYamamoto That is not a relevant link. How is the boxing/unboxing behavior of Nullable<T> possible? is.

– GSerg
Mar 27 at 11:17












3 Answers
3






active

oldest

votes


















54














Nullable types have special "boxing" rules; "boxing" is when a value-type is treated as object, as per your code. Unlike regular value-types, a nullable value-type is boxed either as null (regular null, no type), or as the non-nullable type (the T in T?). So: an int? is boxed as an int, not an int?. Then when you use OfType<int>() on it, you get all the values that are int, which is: the single value you passed in, since it is of type int.






share|improve this answer























  • phew ok thank you for that explanaion. Is that C# basic knowledge?

    – Dr. Snail
    Mar 27 at 8:30






  • 14





    @Dr.Snail "basic" is relative / subjective, and I'd wager that a good percentage of developers never have a need to know that nuance; it is useful context if you're dealing with boxing, though... and technically it isn't really C# knowledge, but rather: .NET knowledge (it would apply to all languages)

    – Marc Gravell
    Mar 27 at 8:31







  • 5





    @Sinatr you can't - the list never contains int? - it only contains int because of the boxing rules on nullable types

    – Marc Gravell
    Mar 27 at 8:34







  • 1





    @KyleJohnson you added a nullable int to the list. Naively, if nullable ints and ints are different things, you'd expect asking for all the ints in the list to return nothing. There are languages that do it that way, but C# has chosen to do it differently.

    – mbrig
    Mar 27 at 21:24






  • 1





    (and yes, my first sentence there is technically wrong. The nullable int never got added to the list. But if you don't know what's happening here, that's what it looks like is happening)

    – mbrig
    Mar 27 at 21:26


















8














A nullable value type is boxed by the following rules:



  • If HasValue returns false, the null reference is produced.

  • If HasValue returns true, a value of the underlying value type T is
    boxed, not the instance of nullable.


In your example second rule has been followed as you have value, e.g.
var i = (object)(int?)123;






share|improve this answer
































    3














    It is a bit late, but beside of Marc's answer to your question, I want to give some additional information about Nullable value types in CLR.



    The CLR has built-in support for nullable value types. This special support is provided for boxing, unboxing, calling GetType, calling interface methods.



    For example, let's check GetType():



    Int32? x = 5;
    Console.WriteLine(x.GetType());


    What you think it will print to the console?
    System.Nullable<Int32? Not, the result is System.Int32.



    Or, let's check boxing, which you noted in your question:



    Int32? n =5;
    Object o = n;
    Console.WriteLine("o's type=0", o.GetType()); // "System.Int32"


    The rule is that:




    When the CLR is boxing a Nullable instance, it checks to see if it
    is null, and if so, the CLR doesn’t actually box anything, and null is
    returned. If the nullable instance is not null, the CLR takes the
    value out of the nullable instance and boxes it. In other words, a
    Nullable with a value of 5 is boxed into a boxed-Int32 with a
    value of 5.




    And, at the end I want to explain how CLR add special support for calling interface methods from Nullable Types.
    Let's take a look to that:



    Int32? n = 5;
    Int32 result = ((IComparable) n).CompareTo(5); // Compiles & runs OK
    Console.WriteLine(result); // 0


    In the preceding code, I’m casting n, a Nullable<Int32>, to IComparable<Int32>, an interface
    type. However, the Nullable<T> type does not implement the IComparable<Int32> interface as
    Int32 does. The C# compiler allows this code to compile anyway.






    share|improve this answer

























    • So, nullables have special treatment by the compiler? I couldn't implement these rules for my own type?

      – Markonius
      2 days ago






    • 1





      @Markonius For example you can not override GetType for your own type to lie about the real type of your object. Because, GetType is non-virtual.

      – Farhad Jabiyev
      2 days ago











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    3 Answers
    3






    active

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    54














    Nullable types have special "boxing" rules; "boxing" is when a value-type is treated as object, as per your code. Unlike regular value-types, a nullable value-type is boxed either as null (regular null, no type), or as the non-nullable type (the T in T?). So: an int? is boxed as an int, not an int?. Then when you use OfType<int>() on it, you get all the values that are int, which is: the single value you passed in, since it is of type int.






    share|improve this answer























    • phew ok thank you for that explanaion. Is that C# basic knowledge?

      – Dr. Snail
      Mar 27 at 8:30






    • 14





      @Dr.Snail "basic" is relative / subjective, and I'd wager that a good percentage of developers never have a need to know that nuance; it is useful context if you're dealing with boxing, though... and technically it isn't really C# knowledge, but rather: .NET knowledge (it would apply to all languages)

      – Marc Gravell
      Mar 27 at 8:31







    • 5





      @Sinatr you can't - the list never contains int? - it only contains int because of the boxing rules on nullable types

      – Marc Gravell
      Mar 27 at 8:34







    • 1





      @KyleJohnson you added a nullable int to the list. Naively, if nullable ints and ints are different things, you'd expect asking for all the ints in the list to return nothing. There are languages that do it that way, but C# has chosen to do it differently.

      – mbrig
      Mar 27 at 21:24






    • 1





      (and yes, my first sentence there is technically wrong. The nullable int never got added to the list. But if you don't know what's happening here, that's what it looks like is happening)

      – mbrig
      Mar 27 at 21:26















    54














    Nullable types have special "boxing" rules; "boxing" is when a value-type is treated as object, as per your code. Unlike regular value-types, a nullable value-type is boxed either as null (regular null, no type), or as the non-nullable type (the T in T?). So: an int? is boxed as an int, not an int?. Then when you use OfType<int>() on it, you get all the values that are int, which is: the single value you passed in, since it is of type int.






    share|improve this answer























    • phew ok thank you for that explanaion. Is that C# basic knowledge?

      – Dr. Snail
      Mar 27 at 8:30






    • 14





      @Dr.Snail "basic" is relative / subjective, and I'd wager that a good percentage of developers never have a need to know that nuance; it is useful context if you're dealing with boxing, though... and technically it isn't really C# knowledge, but rather: .NET knowledge (it would apply to all languages)

      – Marc Gravell
      Mar 27 at 8:31







    • 5





      @Sinatr you can't - the list never contains int? - it only contains int because of the boxing rules on nullable types

      – Marc Gravell
      Mar 27 at 8:34







    • 1





      @KyleJohnson you added a nullable int to the list. Naively, if nullable ints and ints are different things, you'd expect asking for all the ints in the list to return nothing. There are languages that do it that way, but C# has chosen to do it differently.

      – mbrig
      Mar 27 at 21:24






    • 1





      (and yes, my first sentence there is technically wrong. The nullable int never got added to the list. But if you don't know what's happening here, that's what it looks like is happening)

      – mbrig
      Mar 27 at 21:26













    54












    54








    54







    Nullable types have special "boxing" rules; "boxing" is when a value-type is treated as object, as per your code. Unlike regular value-types, a nullable value-type is boxed either as null (regular null, no type), or as the non-nullable type (the T in T?). So: an int? is boxed as an int, not an int?. Then when you use OfType<int>() on it, you get all the values that are int, which is: the single value you passed in, since it is of type int.






    share|improve this answer













    Nullable types have special "boxing" rules; "boxing" is when a value-type is treated as object, as per your code. Unlike regular value-types, a nullable value-type is boxed either as null (regular null, no type), or as the non-nullable type (the T in T?). So: an int? is boxed as an int, not an int?. Then when you use OfType<int>() on it, you get all the values that are int, which is: the single value you passed in, since it is of type int.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Mar 27 at 8:28









    Marc GravellMarc Gravell

    794k19821612565




    794k19821612565












    • phew ok thank you for that explanaion. Is that C# basic knowledge?

      – Dr. Snail
      Mar 27 at 8:30






    • 14





      @Dr.Snail "basic" is relative / subjective, and I'd wager that a good percentage of developers never have a need to know that nuance; it is useful context if you're dealing with boxing, though... and technically it isn't really C# knowledge, but rather: .NET knowledge (it would apply to all languages)

      – Marc Gravell
      Mar 27 at 8:31







    • 5





      @Sinatr you can't - the list never contains int? - it only contains int because of the boxing rules on nullable types

      – Marc Gravell
      Mar 27 at 8:34







    • 1





      @KyleJohnson you added a nullable int to the list. Naively, if nullable ints and ints are different things, you'd expect asking for all the ints in the list to return nothing. There are languages that do it that way, but C# has chosen to do it differently.

      – mbrig
      Mar 27 at 21:24






    • 1





      (and yes, my first sentence there is technically wrong. The nullable int never got added to the list. But if you don't know what's happening here, that's what it looks like is happening)

      – mbrig
      Mar 27 at 21:26

















    • phew ok thank you for that explanaion. Is that C# basic knowledge?

      – Dr. Snail
      Mar 27 at 8:30






    • 14





      @Dr.Snail "basic" is relative / subjective, and I'd wager that a good percentage of developers never have a need to know that nuance; it is useful context if you're dealing with boxing, though... and technically it isn't really C# knowledge, but rather: .NET knowledge (it would apply to all languages)

      – Marc Gravell
      Mar 27 at 8:31







    • 5





      @Sinatr you can't - the list never contains int? - it only contains int because of the boxing rules on nullable types

      – Marc Gravell
      Mar 27 at 8:34







    • 1





      @KyleJohnson you added a nullable int to the list. Naively, if nullable ints and ints are different things, you'd expect asking for all the ints in the list to return nothing. There are languages that do it that way, but C# has chosen to do it differently.

      – mbrig
      Mar 27 at 21:24






    • 1





      (and yes, my first sentence there is technically wrong. The nullable int never got added to the list. But if you don't know what's happening here, that's what it looks like is happening)

      – mbrig
      Mar 27 at 21:26
















    phew ok thank you for that explanaion. Is that C# basic knowledge?

    – Dr. Snail
    Mar 27 at 8:30





    phew ok thank you for that explanaion. Is that C# basic knowledge?

    – Dr. Snail
    Mar 27 at 8:30




    14




    14





    @Dr.Snail "basic" is relative / subjective, and I'd wager that a good percentage of developers never have a need to know that nuance; it is useful context if you're dealing with boxing, though... and technically it isn't really C# knowledge, but rather: .NET knowledge (it would apply to all languages)

    – Marc Gravell
    Mar 27 at 8:31






    @Dr.Snail "basic" is relative / subjective, and I'd wager that a good percentage of developers never have a need to know that nuance; it is useful context if you're dealing with boxing, though... and technically it isn't really C# knowledge, but rather: .NET knowledge (it would apply to all languages)

    – Marc Gravell
    Mar 27 at 8:31





    5




    5





    @Sinatr you can't - the list never contains int? - it only contains int because of the boxing rules on nullable types

    – Marc Gravell
    Mar 27 at 8:34






    @Sinatr you can't - the list never contains int? - it only contains int because of the boxing rules on nullable types

    – Marc Gravell
    Mar 27 at 8:34





    1




    1





    @KyleJohnson you added a nullable int to the list. Naively, if nullable ints and ints are different things, you'd expect asking for all the ints in the list to return nothing. There are languages that do it that way, but C# has chosen to do it differently.

    – mbrig
    Mar 27 at 21:24





    @KyleJohnson you added a nullable int to the list. Naively, if nullable ints and ints are different things, you'd expect asking for all the ints in the list to return nothing. There are languages that do it that way, but C# has chosen to do it differently.

    – mbrig
    Mar 27 at 21:24




    1




    1





    (and yes, my first sentence there is technically wrong. The nullable int never got added to the list. But if you don't know what's happening here, that's what it looks like is happening)

    – mbrig
    Mar 27 at 21:26





    (and yes, my first sentence there is technically wrong. The nullable int never got added to the list. But if you don't know what's happening here, that's what it looks like is happening)

    – mbrig
    Mar 27 at 21:26













    8














    A nullable value type is boxed by the following rules:



    • If HasValue returns false, the null reference is produced.

    • If HasValue returns true, a value of the underlying value type T is
      boxed, not the instance of nullable.


    In your example second rule has been followed as you have value, e.g.
    var i = (object)(int?)123;






    share|improve this answer





























      8














      A nullable value type is boxed by the following rules:



      • If HasValue returns false, the null reference is produced.

      • If HasValue returns true, a value of the underlying value type T is
        boxed, not the instance of nullable.


      In your example second rule has been followed as you have value, e.g.
      var i = (object)(int?)123;






      share|improve this answer



























        8












        8








        8







        A nullable value type is boxed by the following rules:



        • If HasValue returns false, the null reference is produced.

        • If HasValue returns true, a value of the underlying value type T is
          boxed, not the instance of nullable.


        In your example second rule has been followed as you have value, e.g.
        var i = (object)(int?)123;






        share|improve this answer















        A nullable value type is boxed by the following rules:



        • If HasValue returns false, the null reference is produced.

        • If HasValue returns true, a value of the underlying value type T is
          boxed, not the instance of nullable.


        In your example second rule has been followed as you have value, e.g.
        var i = (object)(int?)123;







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 27 at 13:26

























        answered Mar 27 at 9:11









        JohnnyJohnny

        3,8481021




        3,8481021





















            3














            It is a bit late, but beside of Marc's answer to your question, I want to give some additional information about Nullable value types in CLR.



            The CLR has built-in support for nullable value types. This special support is provided for boxing, unboxing, calling GetType, calling interface methods.



            For example, let's check GetType():



            Int32? x = 5;
            Console.WriteLine(x.GetType());


            What you think it will print to the console?
            System.Nullable<Int32? Not, the result is System.Int32.



            Or, let's check boxing, which you noted in your question:



            Int32? n =5;
            Object o = n;
            Console.WriteLine("o's type=0", o.GetType()); // "System.Int32"


            The rule is that:




            When the CLR is boxing a Nullable instance, it checks to see if it
            is null, and if so, the CLR doesn’t actually box anything, and null is
            returned. If the nullable instance is not null, the CLR takes the
            value out of the nullable instance and boxes it. In other words, a
            Nullable with a value of 5 is boxed into a boxed-Int32 with a
            value of 5.




            And, at the end I want to explain how CLR add special support for calling interface methods from Nullable Types.
            Let's take a look to that:



            Int32? n = 5;
            Int32 result = ((IComparable) n).CompareTo(5); // Compiles & runs OK
            Console.WriteLine(result); // 0


            In the preceding code, I’m casting n, a Nullable<Int32>, to IComparable<Int32>, an interface
            type. However, the Nullable<T> type does not implement the IComparable<Int32> interface as
            Int32 does. The C# compiler allows this code to compile anyway.






            share|improve this answer

























            • So, nullables have special treatment by the compiler? I couldn't implement these rules for my own type?

              – Markonius
              2 days ago






            • 1





              @Markonius For example you can not override GetType for your own type to lie about the real type of your object. Because, GetType is non-virtual.

              – Farhad Jabiyev
              2 days ago















            3














            It is a bit late, but beside of Marc's answer to your question, I want to give some additional information about Nullable value types in CLR.



            The CLR has built-in support for nullable value types. This special support is provided for boxing, unboxing, calling GetType, calling interface methods.



            For example, let's check GetType():



            Int32? x = 5;
            Console.WriteLine(x.GetType());


            What you think it will print to the console?
            System.Nullable<Int32? Not, the result is System.Int32.



            Or, let's check boxing, which you noted in your question:



            Int32? n =5;
            Object o = n;
            Console.WriteLine("o's type=0", o.GetType()); // "System.Int32"


            The rule is that:




            When the CLR is boxing a Nullable instance, it checks to see if it
            is null, and if so, the CLR doesn’t actually box anything, and null is
            returned. If the nullable instance is not null, the CLR takes the
            value out of the nullable instance and boxes it. In other words, a
            Nullable with a value of 5 is boxed into a boxed-Int32 with a
            value of 5.




            And, at the end I want to explain how CLR add special support for calling interface methods from Nullable Types.
            Let's take a look to that:



            Int32? n = 5;
            Int32 result = ((IComparable) n).CompareTo(5); // Compiles & runs OK
            Console.WriteLine(result); // 0


            In the preceding code, I’m casting n, a Nullable<Int32>, to IComparable<Int32>, an interface
            type. However, the Nullable<T> type does not implement the IComparable<Int32> interface as
            Int32 does. The C# compiler allows this code to compile anyway.






            share|improve this answer

























            • So, nullables have special treatment by the compiler? I couldn't implement these rules for my own type?

              – Markonius
              2 days ago






            • 1





              @Markonius For example you can not override GetType for your own type to lie about the real type of your object. Because, GetType is non-virtual.

              – Farhad Jabiyev
              2 days ago













            3












            3








            3







            It is a bit late, but beside of Marc's answer to your question, I want to give some additional information about Nullable value types in CLR.



            The CLR has built-in support for nullable value types. This special support is provided for boxing, unboxing, calling GetType, calling interface methods.



            For example, let's check GetType():



            Int32? x = 5;
            Console.WriteLine(x.GetType());


            What you think it will print to the console?
            System.Nullable<Int32? Not, the result is System.Int32.



            Or, let's check boxing, which you noted in your question:



            Int32? n =5;
            Object o = n;
            Console.WriteLine("o's type=0", o.GetType()); // "System.Int32"


            The rule is that:




            When the CLR is boxing a Nullable instance, it checks to see if it
            is null, and if so, the CLR doesn’t actually box anything, and null is
            returned. If the nullable instance is not null, the CLR takes the
            value out of the nullable instance and boxes it. In other words, a
            Nullable with a value of 5 is boxed into a boxed-Int32 with a
            value of 5.




            And, at the end I want to explain how CLR add special support for calling interface methods from Nullable Types.
            Let's take a look to that:



            Int32? n = 5;
            Int32 result = ((IComparable) n).CompareTo(5); // Compiles & runs OK
            Console.WriteLine(result); // 0


            In the preceding code, I’m casting n, a Nullable<Int32>, to IComparable<Int32>, an interface
            type. However, the Nullable<T> type does not implement the IComparable<Int32> interface as
            Int32 does. The C# compiler allows this code to compile anyway.






            share|improve this answer















            It is a bit late, but beside of Marc's answer to your question, I want to give some additional information about Nullable value types in CLR.



            The CLR has built-in support for nullable value types. This special support is provided for boxing, unboxing, calling GetType, calling interface methods.



            For example, let's check GetType():



            Int32? x = 5;
            Console.WriteLine(x.GetType());


            What you think it will print to the console?
            System.Nullable<Int32? Not, the result is System.Int32.



            Or, let's check boxing, which you noted in your question:



            Int32? n =5;
            Object o = n;
            Console.WriteLine("o's type=0", o.GetType()); // "System.Int32"


            The rule is that:




            When the CLR is boxing a Nullable instance, it checks to see if it
            is null, and if so, the CLR doesn’t actually box anything, and null is
            returned. If the nullable instance is not null, the CLR takes the
            value out of the nullable instance and boxes it. In other words, a
            Nullable with a value of 5 is boxed into a boxed-Int32 with a
            value of 5.




            And, at the end I want to explain how CLR add special support for calling interface methods from Nullable Types.
            Let's take a look to that:



            Int32? n = 5;
            Int32 result = ((IComparable) n).CompareTo(5); // Compiles & runs OK
            Console.WriteLine(result); // 0


            In the preceding code, I’m casting n, a Nullable<Int32>, to IComparable<Int32>, an interface
            type. However, the Nullable<T> type does not implement the IComparable<Int32> interface as
            Int32 does. The C# compiler allows this code to compile anyway.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Apr 2 at 21:53

























            answered Apr 2 at 17:59









            Farhad JabiyevFarhad Jabiyev

            19k64279




            19k64279












            • So, nullables have special treatment by the compiler? I couldn't implement these rules for my own type?

              – Markonius
              2 days ago






            • 1





              @Markonius For example you can not override GetType for your own type to lie about the real type of your object. Because, GetType is non-virtual.

              – Farhad Jabiyev
              2 days ago

















            • So, nullables have special treatment by the compiler? I couldn't implement these rules for my own type?

              – Markonius
              2 days ago






            • 1





              @Markonius For example you can not override GetType for your own type to lie about the real type of your object. Because, GetType is non-virtual.

              – Farhad Jabiyev
              2 days ago
















            So, nullables have special treatment by the compiler? I couldn't implement these rules for my own type?

            – Markonius
            2 days ago





            So, nullables have special treatment by the compiler? I couldn't implement these rules for my own type?

            – Markonius
            2 days ago




            1




            1





            @Markonius For example you can not override GetType for your own type to lie about the real type of your object. Because, GetType is non-virtual.

            – Farhad Jabiyev
            2 days ago





            @Markonius For example you can not override GetType for your own type to lie about the real type of your object. Because, GetType is non-virtual.

            – Farhad Jabiyev
            2 days ago

















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