why in c# casting object to objects of type T with same value aren't recognized the same?












4















I am getting into generics but there is something I don't understand Here I have this class;



    class Node<T>{
T value;
Node <T> next,prev;
public Node(T vl)
{
this.value=vl;
this.next=this.prev=null;
}
}


I have a class List and a method Insert and it is working fine :



 class Lista<T>
{
public Node<T> Head, Tail;
public int Cnt;

public Lista()
{
this.Head = this.Tail = null;
this.Cnt = 0;
}



public void Insert(T vl)
{
Node<T> nd = new Node<T>(vl);
if (this.IsEmpty())
this.Head = this.Tail = nd;
else
{
this.Tail.next = nd;
nd.prev = this.Tail;
this.Tail = nd;
}
this.Cnt++;
}


}


but I Have a method FindNode:



 public Node<T> FindNode(T vl)
{
if (this.IsEmpty()) return null;
Node<T> tmp = this.Head;
while (tmp != null && ((object)tmp.VL != (object)vl)) tmp = tmp.next;
return tmp;
}


which finds a Node inside the List now the problem is
if I have a Lista<string> it works fine but if I create a Lista<int>
even if the element in list exists while comparing it is ignoring the same values and going next until it goes null , why does that happen ?



Method isEmpty:



public bool IsEmpty()
{
if (this.Head == null && this.Tail == null) return true;
return false;
}









share|improve this question


















  • 1





    Generics was invented to avoid something like (object)tmp.VL.

    – SeM
    Nov 22 '18 at 17:36











  • You haven't overloaded Equals so how is C# going to know you consider them the same? It can only check whether the variables refer to the same instance

    – Panagiotis Kanavos
    Nov 22 '18 at 17:36













  • @SeM I know right but if I say tmp.VL==vl it says that operand== cannot be used to types T and T

    – Johnny Adams
    Nov 22 '18 at 17:38











  • @PanagiotisKanavos how do I do that could you explain please

    – Johnny Adams
    Nov 22 '18 at 17:38











  • @JohnnyAdams instead of trying to cover up the compilation problem, fix it. Override Equals and ==

    – Panagiotis Kanavos
    Nov 22 '18 at 17:39
















4















I am getting into generics but there is something I don't understand Here I have this class;



    class Node<T>{
T value;
Node <T> next,prev;
public Node(T vl)
{
this.value=vl;
this.next=this.prev=null;
}
}


I have a class List and a method Insert and it is working fine :



 class Lista<T>
{
public Node<T> Head, Tail;
public int Cnt;

public Lista()
{
this.Head = this.Tail = null;
this.Cnt = 0;
}



public void Insert(T vl)
{
Node<T> nd = new Node<T>(vl);
if (this.IsEmpty())
this.Head = this.Tail = nd;
else
{
this.Tail.next = nd;
nd.prev = this.Tail;
this.Tail = nd;
}
this.Cnt++;
}


}


but I Have a method FindNode:



 public Node<T> FindNode(T vl)
{
if (this.IsEmpty()) return null;
Node<T> tmp = this.Head;
while (tmp != null && ((object)tmp.VL != (object)vl)) tmp = tmp.next;
return tmp;
}


which finds a Node inside the List now the problem is
if I have a Lista<string> it works fine but if I create a Lista<int>
even if the element in list exists while comparing it is ignoring the same values and going next until it goes null , why does that happen ?



Method isEmpty:



public bool IsEmpty()
{
if (this.Head == null && this.Tail == null) return true;
return false;
}









share|improve this question


















  • 1





    Generics was invented to avoid something like (object)tmp.VL.

    – SeM
    Nov 22 '18 at 17:36











  • You haven't overloaded Equals so how is C# going to know you consider them the same? It can only check whether the variables refer to the same instance

    – Panagiotis Kanavos
    Nov 22 '18 at 17:36













  • @SeM I know right but if I say tmp.VL==vl it says that operand== cannot be used to types T and T

    – Johnny Adams
    Nov 22 '18 at 17:38











  • @PanagiotisKanavos how do I do that could you explain please

    – Johnny Adams
    Nov 22 '18 at 17:38











  • @JohnnyAdams instead of trying to cover up the compilation problem, fix it. Override Equals and ==

    – Panagiotis Kanavos
    Nov 22 '18 at 17:39














4












4








4








I am getting into generics but there is something I don't understand Here I have this class;



    class Node<T>{
T value;
Node <T> next,prev;
public Node(T vl)
{
this.value=vl;
this.next=this.prev=null;
}
}


I have a class List and a method Insert and it is working fine :



 class Lista<T>
{
public Node<T> Head, Tail;
public int Cnt;

public Lista()
{
this.Head = this.Tail = null;
this.Cnt = 0;
}



public void Insert(T vl)
{
Node<T> nd = new Node<T>(vl);
if (this.IsEmpty())
this.Head = this.Tail = nd;
else
{
this.Tail.next = nd;
nd.prev = this.Tail;
this.Tail = nd;
}
this.Cnt++;
}


}


but I Have a method FindNode:



 public Node<T> FindNode(T vl)
{
if (this.IsEmpty()) return null;
Node<T> tmp = this.Head;
while (tmp != null && ((object)tmp.VL != (object)vl)) tmp = tmp.next;
return tmp;
}


which finds a Node inside the List now the problem is
if I have a Lista<string> it works fine but if I create a Lista<int>
even if the element in list exists while comparing it is ignoring the same values and going next until it goes null , why does that happen ?



Method isEmpty:



public bool IsEmpty()
{
if (this.Head == null && this.Tail == null) return true;
return false;
}









share|improve this question














I am getting into generics but there is something I don't understand Here I have this class;



    class Node<T>{
T value;
Node <T> next,prev;
public Node(T vl)
{
this.value=vl;
this.next=this.prev=null;
}
}


I have a class List and a method Insert and it is working fine :



 class Lista<T>
{
public Node<T> Head, Tail;
public int Cnt;

public Lista()
{
this.Head = this.Tail = null;
this.Cnt = 0;
}



public void Insert(T vl)
{
Node<T> nd = new Node<T>(vl);
if (this.IsEmpty())
this.Head = this.Tail = nd;
else
{
this.Tail.next = nd;
nd.prev = this.Tail;
this.Tail = nd;
}
this.Cnt++;
}


}


but I Have a method FindNode:



 public Node<T> FindNode(T vl)
{
if (this.IsEmpty()) return null;
Node<T> tmp = this.Head;
while (tmp != null && ((object)tmp.VL != (object)vl)) tmp = tmp.next;
return tmp;
}


which finds a Node inside the List now the problem is
if I have a Lista<string> it works fine but if I create a Lista<int>
even if the element in list exists while comparing it is ignoring the same values and going next until it goes null , why does that happen ?



Method isEmpty:



public bool IsEmpty()
{
if (this.Head == null && this.Tail == null) return true;
return false;
}






c# generics






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 22 '18 at 17:33









Johnny AdamsJohnny Adams

256




256








  • 1





    Generics was invented to avoid something like (object)tmp.VL.

    – SeM
    Nov 22 '18 at 17:36











  • You haven't overloaded Equals so how is C# going to know you consider them the same? It can only check whether the variables refer to the same instance

    – Panagiotis Kanavos
    Nov 22 '18 at 17:36













  • @SeM I know right but if I say tmp.VL==vl it says that operand== cannot be used to types T and T

    – Johnny Adams
    Nov 22 '18 at 17:38











  • @PanagiotisKanavos how do I do that could you explain please

    – Johnny Adams
    Nov 22 '18 at 17:38











  • @JohnnyAdams instead of trying to cover up the compilation problem, fix it. Override Equals and ==

    – Panagiotis Kanavos
    Nov 22 '18 at 17:39














  • 1





    Generics was invented to avoid something like (object)tmp.VL.

    – SeM
    Nov 22 '18 at 17:36











  • You haven't overloaded Equals so how is C# going to know you consider them the same? It can only check whether the variables refer to the same instance

    – Panagiotis Kanavos
    Nov 22 '18 at 17:36













  • @SeM I know right but if I say tmp.VL==vl it says that operand== cannot be used to types T and T

    – Johnny Adams
    Nov 22 '18 at 17:38











  • @PanagiotisKanavos how do I do that could you explain please

    – Johnny Adams
    Nov 22 '18 at 17:38











  • @JohnnyAdams instead of trying to cover up the compilation problem, fix it. Override Equals and ==

    – Panagiotis Kanavos
    Nov 22 '18 at 17:39








1




1





Generics was invented to avoid something like (object)tmp.VL.

– SeM
Nov 22 '18 at 17:36





Generics was invented to avoid something like (object)tmp.VL.

– SeM
Nov 22 '18 at 17:36













You haven't overloaded Equals so how is C# going to know you consider them the same? It can only check whether the variables refer to the same instance

– Panagiotis Kanavos
Nov 22 '18 at 17:36







You haven't overloaded Equals so how is C# going to know you consider them the same? It can only check whether the variables refer to the same instance

– Panagiotis Kanavos
Nov 22 '18 at 17:36















@SeM I know right but if I say tmp.VL==vl it says that operand== cannot be used to types T and T

– Johnny Adams
Nov 22 '18 at 17:38





@SeM I know right but if I say tmp.VL==vl it says that operand== cannot be used to types T and T

– Johnny Adams
Nov 22 '18 at 17:38













@PanagiotisKanavos how do I do that could you explain please

– Johnny Adams
Nov 22 '18 at 17:38





@PanagiotisKanavos how do I do that could you explain please

– Johnny Adams
Nov 22 '18 at 17:38













@JohnnyAdams instead of trying to cover up the compilation problem, fix it. Override Equals and ==

– Panagiotis Kanavos
Nov 22 '18 at 17:39





@JohnnyAdams instead of trying to cover up the compilation problem, fix it. Override Equals and ==

– Panagiotis Kanavos
Nov 22 '18 at 17:39












1 Answer
1






active

oldest

votes


















4














The problem is, that for valuetypes like int in your example you are boxing them by casting to object.
Per default, the == operator does ReferenceEquality and this means that it will never return true in your case because the boxes are never the same.



Easiest was to fix this would be to turn this line



while (tmp != null && ((object)tmp.VL != (object)vl)) tmp = tmp.next;


into this



 while (tmp != null && !Equals(tmp.VL , vl)) tmp = tmp.next;


This will use the default equality comparer for a certain type and makes your code work as intended.



You could go further by using a generic constraint declared for your class like so:



class Lista<T> where T : IEquateable<T>


and change the same line like this



while (tmp != null && (!tmp.VL.Equals(vl)) tmp = tmp.next;


Or make it possible to inject an IEqualityComparer<T> into your class



  class Lista<T>
{
IEqualityComparer<T> _comparer;
public Lista(IEqualityComparer<T> comparer = null)
{
_comparer = comparer ?? EuqlityComparer<T>.Default;
}

public Node<T> FindNode(T vl)
{
if (this.IsEmpty()) return null;
Node<T> tmp = this.Head;
while (tmp != null && !_comparer.Equals(tmp.VL, vl) tmp = tmp.next;
return tmp;
}
}





share|improve this answer


























  • thanks this was simple and worked like charm , but from the both methods you suggested which one would you recommend to use

    – Johnny Adams
    Nov 22 '18 at 17:47






  • 1





    @JohnnyAdams the last one, using the IEqualityComparer is usually common practice. But if you dont need it, the first one is just fine.

    – CSharpie
    Nov 22 '18 at 17:47













  • okay thanks for your answer

    – Johnny Adams
    Nov 22 '18 at 17:49











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4














The problem is, that for valuetypes like int in your example you are boxing them by casting to object.
Per default, the == operator does ReferenceEquality and this means that it will never return true in your case because the boxes are never the same.



Easiest was to fix this would be to turn this line



while (tmp != null && ((object)tmp.VL != (object)vl)) tmp = tmp.next;


into this



 while (tmp != null && !Equals(tmp.VL , vl)) tmp = tmp.next;


This will use the default equality comparer for a certain type and makes your code work as intended.



You could go further by using a generic constraint declared for your class like so:



class Lista<T> where T : IEquateable<T>


and change the same line like this



while (tmp != null && (!tmp.VL.Equals(vl)) tmp = tmp.next;


Or make it possible to inject an IEqualityComparer<T> into your class



  class Lista<T>
{
IEqualityComparer<T> _comparer;
public Lista(IEqualityComparer<T> comparer = null)
{
_comparer = comparer ?? EuqlityComparer<T>.Default;
}

public Node<T> FindNode(T vl)
{
if (this.IsEmpty()) return null;
Node<T> tmp = this.Head;
while (tmp != null && !_comparer.Equals(tmp.VL, vl) tmp = tmp.next;
return tmp;
}
}





share|improve this answer


























  • thanks this was simple and worked like charm , but from the both methods you suggested which one would you recommend to use

    – Johnny Adams
    Nov 22 '18 at 17:47






  • 1





    @JohnnyAdams the last one, using the IEqualityComparer is usually common practice. But if you dont need it, the first one is just fine.

    – CSharpie
    Nov 22 '18 at 17:47













  • okay thanks for your answer

    – Johnny Adams
    Nov 22 '18 at 17:49
















4














The problem is, that for valuetypes like int in your example you are boxing them by casting to object.
Per default, the == operator does ReferenceEquality and this means that it will never return true in your case because the boxes are never the same.



Easiest was to fix this would be to turn this line



while (tmp != null && ((object)tmp.VL != (object)vl)) tmp = tmp.next;


into this



 while (tmp != null && !Equals(tmp.VL , vl)) tmp = tmp.next;


This will use the default equality comparer for a certain type and makes your code work as intended.



You could go further by using a generic constraint declared for your class like so:



class Lista<T> where T : IEquateable<T>


and change the same line like this



while (tmp != null && (!tmp.VL.Equals(vl)) tmp = tmp.next;


Or make it possible to inject an IEqualityComparer<T> into your class



  class Lista<T>
{
IEqualityComparer<T> _comparer;
public Lista(IEqualityComparer<T> comparer = null)
{
_comparer = comparer ?? EuqlityComparer<T>.Default;
}

public Node<T> FindNode(T vl)
{
if (this.IsEmpty()) return null;
Node<T> tmp = this.Head;
while (tmp != null && !_comparer.Equals(tmp.VL, vl) tmp = tmp.next;
return tmp;
}
}





share|improve this answer


























  • thanks this was simple and worked like charm , but from the both methods you suggested which one would you recommend to use

    – Johnny Adams
    Nov 22 '18 at 17:47






  • 1





    @JohnnyAdams the last one, using the IEqualityComparer is usually common practice. But if you dont need it, the first one is just fine.

    – CSharpie
    Nov 22 '18 at 17:47













  • okay thanks for your answer

    – Johnny Adams
    Nov 22 '18 at 17:49














4












4








4







The problem is, that for valuetypes like int in your example you are boxing them by casting to object.
Per default, the == operator does ReferenceEquality and this means that it will never return true in your case because the boxes are never the same.



Easiest was to fix this would be to turn this line



while (tmp != null && ((object)tmp.VL != (object)vl)) tmp = tmp.next;


into this



 while (tmp != null && !Equals(tmp.VL , vl)) tmp = tmp.next;


This will use the default equality comparer for a certain type and makes your code work as intended.



You could go further by using a generic constraint declared for your class like so:



class Lista<T> where T : IEquateable<T>


and change the same line like this



while (tmp != null && (!tmp.VL.Equals(vl)) tmp = tmp.next;


Or make it possible to inject an IEqualityComparer<T> into your class



  class Lista<T>
{
IEqualityComparer<T> _comparer;
public Lista(IEqualityComparer<T> comparer = null)
{
_comparer = comparer ?? EuqlityComparer<T>.Default;
}

public Node<T> FindNode(T vl)
{
if (this.IsEmpty()) return null;
Node<T> tmp = this.Head;
while (tmp != null && !_comparer.Equals(tmp.VL, vl) tmp = tmp.next;
return tmp;
}
}





share|improve this answer















The problem is, that for valuetypes like int in your example you are boxing them by casting to object.
Per default, the == operator does ReferenceEquality and this means that it will never return true in your case because the boxes are never the same.



Easiest was to fix this would be to turn this line



while (tmp != null && ((object)tmp.VL != (object)vl)) tmp = tmp.next;


into this



 while (tmp != null && !Equals(tmp.VL , vl)) tmp = tmp.next;


This will use the default equality comparer for a certain type and makes your code work as intended.



You could go further by using a generic constraint declared for your class like so:



class Lista<T> where T : IEquateable<T>


and change the same line like this



while (tmp != null && (!tmp.VL.Equals(vl)) tmp = tmp.next;


Or make it possible to inject an IEqualityComparer<T> into your class



  class Lista<T>
{
IEqualityComparer<T> _comparer;
public Lista(IEqualityComparer<T> comparer = null)
{
_comparer = comparer ?? EuqlityComparer<T>.Default;
}

public Node<T> FindNode(T vl)
{
if (this.IsEmpty()) return null;
Node<T> tmp = this.Head;
while (tmp != null && !_comparer.Equals(tmp.VL, vl) tmp = tmp.next;
return tmp;
}
}






share|improve this answer














share|improve this answer



share|improve this answer








edited Jan 24 at 6:46

























answered Nov 22 '18 at 17:38









CSharpieCSharpie

6,31023354




6,31023354













  • thanks this was simple and worked like charm , but from the both methods you suggested which one would you recommend to use

    – Johnny Adams
    Nov 22 '18 at 17:47






  • 1





    @JohnnyAdams the last one, using the IEqualityComparer is usually common practice. But if you dont need it, the first one is just fine.

    – CSharpie
    Nov 22 '18 at 17:47













  • okay thanks for your answer

    – Johnny Adams
    Nov 22 '18 at 17:49



















  • thanks this was simple and worked like charm , but from the both methods you suggested which one would you recommend to use

    – Johnny Adams
    Nov 22 '18 at 17:47






  • 1





    @JohnnyAdams the last one, using the IEqualityComparer is usually common practice. But if you dont need it, the first one is just fine.

    – CSharpie
    Nov 22 '18 at 17:47













  • okay thanks for your answer

    – Johnny Adams
    Nov 22 '18 at 17:49

















thanks this was simple and worked like charm , but from the both methods you suggested which one would you recommend to use

– Johnny Adams
Nov 22 '18 at 17:47





thanks this was simple and worked like charm , but from the both methods you suggested which one would you recommend to use

– Johnny Adams
Nov 22 '18 at 17:47




1




1





@JohnnyAdams the last one, using the IEqualityComparer is usually common practice. But if you dont need it, the first one is just fine.

– CSharpie
Nov 22 '18 at 17:47







@JohnnyAdams the last one, using the IEqualityComparer is usually common practice. But if you dont need it, the first one is just fine.

– CSharpie
Nov 22 '18 at 17:47















okay thanks for your answer

– Johnny Adams
Nov 22 '18 at 17:49





okay thanks for your answer

– Johnny Adams
Nov 22 '18 at 17:49




















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