Relation between List and IEnumerable open type

Is there any relationship between the open types List<> and IEnumerable<>?

Example:

var type1 = typeof(List<>);

var type2 = typeof(IEnumerable<>);



//return false

type2.IsAssignableFrom(type1);

Is there any method to check the relationship between two open type, or the relationship only exist on closed type?

edited Nov 20 '18 at 9:31

marsze

4,51731640

asked Nov 20 '18 at 7:04

YonF

333112

Relevant: When does Type.FullName return null?
– John Wu
Nov 20 '18 at 7:16

1

Note that each generic parameter is unique. That means List<T(from List<T>)> implement IEnumerable<T(from List<T>)> but not IEnumerable<T(from IEnumerable<T>)>.
– PetSerAl
Nov 20 '18 at 7:22

4

What are you actually trying to accomplish here? If we knew what the initial problem you were trying to solve was, we may be able to offer other solutions that don't rely on comparing open generic types.
– Damien_The_Unbeliever
Nov 20 '18 at 7:59

Sorry for the drastic edit, but I think only the problem expressed in the first 3 lines is the actually interesting part here. What you were trying to do with the rest of the code was very unclear and you would have had to explain that further as @Damien_The_Unbeliever stated.
– marsze
Nov 20 '18 at 9:34

@PetSerAl How to understand each generic parameter is unique? Do you mean type int in List<int> and IEnumerable<int> is "diffierent"?
– YonF
Nov 21 '18 at 3:12

|
show 2 more comments

Is there any relationship between the open types List<> and IEnumerable<>?

Example:

var type1 = typeof(List<>);

var type2 = typeof(IEnumerable<>);



//return false

type2.IsAssignableFrom(type1);

Is there any method to check the relationship between two open type, or the relationship only exist on closed type?

edited Nov 20 '18 at 9:31

marsze

4,51731640

asked Nov 20 '18 at 7:04

YonF

333112

Relevant: When does Type.FullName return null?
– John Wu
Nov 20 '18 at 7:16

1

Note that each generic parameter is unique. That means List<T(from List<T>)> implement IEnumerable<T(from List<T>)> but not IEnumerable<T(from IEnumerable<T>)>.
– PetSerAl
Nov 20 '18 at 7:22

4

What are you actually trying to accomplish here? If we knew what the initial problem you were trying to solve was, we may be able to offer other solutions that don't rely on comparing open generic types.
– Damien_The_Unbeliever
Nov 20 '18 at 7:59

Sorry for the drastic edit, but I think only the problem expressed in the first 3 lines is the actually interesting part here. What you were trying to do with the rest of the code was very unclear and you would have had to explain that further as @Damien_The_Unbeliever stated.
– marsze
Nov 20 '18 at 9:34

@PetSerAl How to understand each generic parameter is unique? Do you mean type int in List<int> and IEnumerable<int> is "diffierent"?
– YonF
Nov 21 '18 at 3:12

|
show 2 more comments

Is there any relationship between the open types List<> and IEnumerable<>?

Example:

var type1 = typeof(List<>);

var type2 = typeof(IEnumerable<>);



//return false

type2.IsAssignableFrom(type1);

Is there any method to check the relationship between two open type, or the relationship only exist on closed type?

edited Nov 20 '18 at 9:31

marsze

4,51731640

asked Nov 20 '18 at 7:04

YonF

333112

Is there any relationship between the open types List<> and IEnumerable<>?

Example:

var type1 = typeof(List<>);

var type2 = typeof(IEnumerable<>);



//return false

type2.IsAssignableFrom(type1);

Is there any method to check the relationship between two open type, or the relationship only exist on closed type?

c# .net

edited Nov 20 '18 at 9:31

marsze

4,51731640

asked Nov 20 '18 at 7:04

YonF

333112

edited Nov 20 '18 at 9:31

marsze

4,51731640

asked Nov 20 '18 at 7:04

YonF

333112

edited Nov 20 '18 at 9:31

marsze

4,51731640

edited Nov 20 '18 at 9:31

marsze

4,51731640

edited Nov 20 '18 at 9:31

marsze

4,51731640

asked Nov 20 '18 at 7:04

YonF

333112

asked Nov 20 '18 at 7:04

YonF

333112

asked Nov 20 '18 at 7:04

YonF

333112

Relevant: When does Type.FullName return null?
– John Wu
Nov 20 '18 at 7:16

1

Note that each generic parameter is unique. That means List<T(from List<T>)> implement IEnumerable<T(from List<T>)> but not IEnumerable<T(from IEnumerable<T>)>.
– PetSerAl
Nov 20 '18 at 7:22

4

What are you actually trying to accomplish here? If we knew what the initial problem you were trying to solve was, we may be able to offer other solutions that don't rely on comparing open generic types.
– Damien_The_Unbeliever
Nov 20 '18 at 7:59

Sorry for the drastic edit, but I think only the problem expressed in the first 3 lines is the actually interesting part here. What you were trying to do with the rest of the code was very unclear and you would have had to explain that further as @Damien_The_Unbeliever stated.
– marsze
Nov 20 '18 at 9:34

@PetSerAl How to understand each generic parameter is unique? Do you mean type int in List<int> and IEnumerable<int> is "diffierent"?
– YonF
Nov 21 '18 at 3:12

|
show 2 more comments

Relevant: When does Type.FullName return null?
– John Wu
Nov 20 '18 at 7:16

1

Note that each generic parameter is unique. That means List<T(from List<T>)> implement IEnumerable<T(from List<T>)> but not IEnumerable<T(from IEnumerable<T>)>.
– PetSerAl
Nov 20 '18 at 7:22

4

What are you actually trying to accomplish here? If we knew what the initial problem you were trying to solve was, we may be able to offer other solutions that don't rely on comparing open generic types.
– Damien_The_Unbeliever
Nov 20 '18 at 7:59

Sorry for the drastic edit, but I think only the problem expressed in the first 3 lines is the actually interesting part here. What you were trying to do with the rest of the code was very unclear and you would have had to explain that further as @Damien_The_Unbeliever stated.
– marsze
Nov 20 '18 at 9:34

@PetSerAl How to understand each generic parameter is unique? Do you mean type int in List<int> and IEnumerable<int> is "diffierent"?
– YonF
Nov 21 '18 at 3:12

Relevant: When does Type.FullName return null?
– John Wu
Nov 20 '18 at 7:16

Note that each generic parameter is unique. That means List<T(from List<T>)> implement IEnumerable<T(from List<T>)> but not IEnumerable<T(from IEnumerable<T>)>.
– PetSerAl
Nov 20 '18 at 7:22

What are you actually trying to accomplish here? If we knew what the initial problem you were trying to solve was, we may be able to offer other solutions that don't rely on comparing open generic types.
– Damien_The_Unbeliever
Nov 20 '18 at 7:59

Sorry for the drastic edit, but I think only the problem expressed in the first 3 lines is the actually interesting part here. What you were trying to do with the rest of the code was very unclear and you would have had to explain that further as @Damien_The_Unbeliever stated.
– marsze
Nov 20 '18 at 9:34

@PetSerAl How to understand each generic parameter is unique? Do you mean type int in List<int> and IEnumerable<int> is "diffierent"?
– YonF
Nov 21 '18 at 3:12

|
show 2 more comments

2 Answers
2

active

oldest

votes

Although John Wu wrote up a nice answer about the differences between a type definition and actual types I don't think it fully answers the problem/question asked in the OP.

Is there any method to check the relationship between two open type,
or the relationship only exist on closed type?

First of all, the relationship always exists; every List<> is always an IEnumerable<>, as you can see in the definition of the List<T> type:

public class List<T> : IList<T>, ICollection<T>, IEnumerable<T>, ...

But this doesn't answer your second question, if there is a way to check if such a relationship exists between two types. You would think that the IsAssignableFrom method can be used to check for existence of a relationship between the type open types, but you can't. Why? Lets find out in the documentation of the IsAssignableFrom function:

Type.IsAssignableFrom (Type c) Returns true if any of the following conditions is true:

c and the current instance represent the same type.

c is derived either directly or indirectly from the current
instance. c is derived directly from the current instance if it
inherits from the current instance; c is derived indirectly from the
current instance if it inherits from a succession of one or more
classes that inherit from the current instance.

The current instance is an interface that c implements.

c is a generic type parameter, and the current instance represents
one of the constraints of c.

and false if none of these conditions are true, or if c is null.

In your case, none of the above mentioned conditions will result in true as: (1) they are not the same type. (2) They cannot be derived from each other as they are open types: their generic type parameters are unknown (unknown != unknown). The List<> implements the closed IEnumerable<T> type and not the open IEnumerable<> type (3). They aren't generic type parameters (4).

To find out if two generic types have a relationship, you would need to inspect their type definitions (and their nested interfaces/base types) and validate that they have a relationship:

public static bool IsAssignableToOpenGenericType(Type givenType, Type genericType)

{

    var interfaceTypes = givenType.GetInterfaces();



    foreach (var it in interfaceTypes)

    {

        if (it.IsGenericType && it.GetGenericTypeDefinition() == genericType)

            return true;

    }



    if (givenType.IsGenericType && givenType.GetGenericTypeDefinition() == genericType)

        return true;



    Type baseType = givenType.BaseType;

    if (baseType == null) return false;



    return IsAssignableToGenericType(baseType, genericType);

}

(source)

Will result in:

    var typ1 = typeof(List<>);

    var typ2 = typeof(IEnumerable<>);



    // true, List<>'s type definition contains an IEnumerable<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ1, typ2));

    // false, IEnumerable<>'s type definition does not contain List<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ2, typ1));

edited Nov 20 '18 at 10:39

answered Nov 20 '18 at 9:11

Jevgeni Geurtsen

2,61141232

add a comment |

List<> and IEnumerable<> are not types; they are type definitions. As such it doesn't really make sense to ask if one is assignable to the other. Neither can be assigned to. You can't declare a variable List<> a = null, for example-- you will get a compilation error "Unexpected use of an unbound generic name."

A type definition becomes a generic type when you specify the type parameter. At that point, it is a type and can be assigned to. So for example List<string> can be assigned to IEnumerable<string>.

If you have a type definition in mind and you want to do a type-compatibility check, just use <object> (or a suitable type if there is a type constraint) instead of <>:

var type1 = typeof(List<object>);

var type2 = typeof(IEnumerable<object>);



//returns true

type2.IsAssignableFrom(type1);

edited Nov 20 '18 at 11:46

vaxquis

7,51453757

answered Nov 20 '18 at 8:42

John Wu

29.4k42652

Thanks for your answer. It is a brief check method. I can understand type2.IsAssignableFrom(type1) behavior. But do you mean two open generic type are independent or no relationship between them before they become closed type or bounded?
– YonF
Nov 21 '18 at 5:32

Can you be specific about what you mean when you say "related?" Of course they have certain relationships, and not others.
– John Wu
Nov 21 '18 at 6:38

A relationship such as inheritance, implimentation but not inheritance and implimentation. But I am not sure whether any kind relatioinship exist between open generic type themselves. @Jevgeni Geurtsen think the special relationship list AssignableFrom exist.
– YonF
Nov 21 '18 at 7:07

Believe it or not, generic type definitions do not have any implementation. The implementation is compiled only when you specify the type parameter, and only for that type (although implementations for reference types can be re-used because all pointers are the same size, but that is just an implementation detail). So you see to ask whether List<> implements IEnumerable<> doesn't make much sense because it has no implementation to begin with.
– John Wu
Nov 21 '18 at 7:15

@Jevgeni Geurtsen think the special relationship list AssignableFrom exist. Oh but it is most definitely there in the framework code, so the relation exists, but as long as it doesn't have a concrete implementation, its useless, as John Wu pointed out.
– Jevgeni Geurtsen
Nov 21 '18 at 8:16

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Although John Wu wrote up a nice answer about the differences between a type definition and actual types I don't think it fully answers the problem/question asked in the OP.

Is there any method to check the relationship between two open type,
or the relationship only exist on closed type?

First of all, the relationship always exists; every List<> is always an IEnumerable<>, as you can see in the definition of the List<T> type:

public class List<T> : IList<T>, ICollection<T>, IEnumerable<T>, ...

Type.IsAssignableFrom (Type c) Returns true if any of the following conditions is true:

c and the current instance represent the same type.

c is derived either directly or indirectly from the current
instance. c is derived directly from the current instance if it
inherits from the current instance; c is derived indirectly from the
current instance if it inherits from a succession of one or more
classes that inherit from the current instance.

The current instance is an interface that c implements.

c is a generic type parameter, and the current instance represents
one of the constraints of c.

and false if none of these conditions are true, or if c is null.

To find out if two generic types have a relationship, you would need to inspect their type definitions (and their nested interfaces/base types) and validate that they have a relationship:

public static bool IsAssignableToOpenGenericType(Type givenType, Type genericType)

{

    var interfaceTypes = givenType.GetInterfaces();



    foreach (var it in interfaceTypes)

    {

        if (it.IsGenericType && it.GetGenericTypeDefinition() == genericType)

            return true;

    }



    if (givenType.IsGenericType && givenType.GetGenericTypeDefinition() == genericType)

        return true;



    Type baseType = givenType.BaseType;

    if (baseType == null) return false;



    return IsAssignableToGenericType(baseType, genericType);

}

(source)

Will result in:

    var typ1 = typeof(List<>);

    var typ2 = typeof(IEnumerable<>);



    // true, List<>'s type definition contains an IEnumerable<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ1, typ2));

    // false, IEnumerable<>'s type definition does not contain List<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ2, typ1));

edited Nov 20 '18 at 10:39

answered Nov 20 '18 at 9:11

Jevgeni Geurtsen

2,61141232

add a comment |

Although John Wu wrote up a nice answer about the differences between a type definition and actual types I don't think it fully answers the problem/question asked in the OP.

Is there any method to check the relationship between two open type,
or the relationship only exist on closed type?

First of all, the relationship always exists; every List<> is always an IEnumerable<>, as you can see in the definition of the List<T> type:

public class List<T> : IList<T>, ICollection<T>, IEnumerable<T>, ...

Type.IsAssignableFrom (Type c) Returns true if any of the following conditions is true:

c and the current instance represent the same type.

c is derived either directly or indirectly from the current
instance. c is derived directly from the current instance if it
inherits from the current instance; c is derived indirectly from the
current instance if it inherits from a succession of one or more
classes that inherit from the current instance.

The current instance is an interface that c implements.

c is a generic type parameter, and the current instance represents
one of the constraints of c.

and false if none of these conditions are true, or if c is null.

To find out if two generic types have a relationship, you would need to inspect their type definitions (and their nested interfaces/base types) and validate that they have a relationship:

public static bool IsAssignableToOpenGenericType(Type givenType, Type genericType)

{

    var interfaceTypes = givenType.GetInterfaces();



    foreach (var it in interfaceTypes)

    {

        if (it.IsGenericType && it.GetGenericTypeDefinition() == genericType)

            return true;

    }



    if (givenType.IsGenericType && givenType.GetGenericTypeDefinition() == genericType)

        return true;



    Type baseType = givenType.BaseType;

    if (baseType == null) return false;



    return IsAssignableToGenericType(baseType, genericType);

}

(source)

Will result in:

    var typ1 = typeof(List<>);

    var typ2 = typeof(IEnumerable<>);



    // true, List<>'s type definition contains an IEnumerable<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ1, typ2));

    // false, IEnumerable<>'s type definition does not contain List<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ2, typ1));

edited Nov 20 '18 at 10:39

answered Nov 20 '18 at 9:11

Jevgeni Geurtsen

2,61141232

add a comment |

Although John Wu wrote up a nice answer about the differences between a type definition and actual types I don't think it fully answers the problem/question asked in the OP.

Is there any method to check the relationship between two open type,
or the relationship only exist on closed type?

First of all, the relationship always exists; every List<> is always an IEnumerable<>, as you can see in the definition of the List<T> type:

public class List<T> : IList<T>, ICollection<T>, IEnumerable<T>, ...

Type.IsAssignableFrom (Type c) Returns true if any of the following conditions is true:

c and the current instance represent the same type.

c is derived either directly or indirectly from the current
instance. c is derived directly from the current instance if it
inherits from the current instance; c is derived indirectly from the
current instance if it inherits from a succession of one or more
classes that inherit from the current instance.

The current instance is an interface that c implements.

c is a generic type parameter, and the current instance represents
one of the constraints of c.

and false if none of these conditions are true, or if c is null.

To find out if two generic types have a relationship, you would need to inspect their type definitions (and their nested interfaces/base types) and validate that they have a relationship:

public static bool IsAssignableToOpenGenericType(Type givenType, Type genericType)

{

    var interfaceTypes = givenType.GetInterfaces();



    foreach (var it in interfaceTypes)

    {

        if (it.IsGenericType && it.GetGenericTypeDefinition() == genericType)

            return true;

    }



    if (givenType.IsGenericType && givenType.GetGenericTypeDefinition() == genericType)

        return true;



    Type baseType = givenType.BaseType;

    if (baseType == null) return false;



    return IsAssignableToGenericType(baseType, genericType);

}

(source)

Will result in:

    var typ1 = typeof(List<>);

    var typ2 = typeof(IEnumerable<>);



    // true, List<>'s type definition contains an IEnumerable<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ1, typ2));

    // false, IEnumerable<>'s type definition does not contain List<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ2, typ1));

edited Nov 20 '18 at 10:39

answered Nov 20 '18 at 9:11

Jevgeni Geurtsen

2,61141232

Although John Wu wrote up a nice answer about the differences between a type definition and actual types I don't think it fully answers the problem/question asked in the OP.

Is there any method to check the relationship between two open type,
or the relationship only exist on closed type?

First of all, the relationship always exists; every List<> is always an IEnumerable<>, as you can see in the definition of the List<T> type:

public class List<T> : IList<T>, ICollection<T>, IEnumerable<T>, ...

Type.IsAssignableFrom (Type c) Returns true if any of the following conditions is true:

c and the current instance represent the same type.

c is derived either directly or indirectly from the current
instance. c is derived directly from the current instance if it
inherits from the current instance; c is derived indirectly from the
current instance if it inherits from a succession of one or more
classes that inherit from the current instance.

The current instance is an interface that c implements.

c is a generic type parameter, and the current instance represents
one of the constraints of c.

and false if none of these conditions are true, or if c is null.

To find out if two generic types have a relationship, you would need to inspect their type definitions (and their nested interfaces/base types) and validate that they have a relationship:

public static bool IsAssignableToOpenGenericType(Type givenType, Type genericType)

{

    var interfaceTypes = givenType.GetInterfaces();



    foreach (var it in interfaceTypes)

    {

        if (it.IsGenericType && it.GetGenericTypeDefinition() == genericType)

            return true;

    }



    if (givenType.IsGenericType && givenType.GetGenericTypeDefinition() == genericType)

        return true;



    Type baseType = givenType.BaseType;

    if (baseType == null) return false;



    return IsAssignableToGenericType(baseType, genericType);

}

(source)

Will result in:

    var typ1 = typeof(List<>);

    var typ2 = typeof(IEnumerable<>);



    // true, List<>'s type definition contains an IEnumerable<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ1, typ2));

    // false, IEnumerable<>'s type definition does not contain List<>

    Console.WriteLine(IsAssignableToOpenGenericType(typ2, typ1));

edited Nov 20 '18 at 10:39

answered Nov 20 '18 at 9:11

Jevgeni Geurtsen

2,61141232

edited Nov 20 '18 at 10:39

answered Nov 20 '18 at 9:11

Jevgeni Geurtsen

2,61141232

answered Nov 20 '18 at 9:11

Jevgeni Geurtsen

2,61141232

answered Nov 20 '18 at 9:11

Jevgeni Geurtsen

2,61141232

add a comment |

If you have a type definition in mind and you want to do a type-compatibility check, just use <object> (or a suitable type if there is a type constraint) instead of <>:

var type1 = typeof(List<object>);

var type2 = typeof(IEnumerable<object>);



//returns true

type2.IsAssignableFrom(type1);

edited Nov 20 '18 at 11:46

vaxquis

7,51453757

answered Nov 20 '18 at 8:42

John Wu

29.4k42652

Thanks for your answer. It is a brief check method. I can understand type2.IsAssignableFrom(type1) behavior. But do you mean two open generic type are independent or no relationship between them before they become closed type or bounded?
– YonF
Nov 21 '18 at 5:32

Can you be specific about what you mean when you say "related?" Of course they have certain relationships, and not others.
– John Wu
Nov 21 '18 at 6:38

A relationship such as inheritance, implimentation but not inheritance and implimentation. But I am not sure whether any kind relatioinship exist between open generic type themselves. @Jevgeni Geurtsen think the special relationship list AssignableFrom exist.
– YonF
Nov 21 '18 at 7:07

Believe it or not, generic type definitions do not have any implementation. The implementation is compiled only when you specify the type parameter, and only for that type (although implementations for reference types can be re-used because all pointers are the same size, but that is just an implementation detail). So you see to ask whether List<> implements IEnumerable<> doesn't make much sense because it has no implementation to begin with.
– John Wu
Nov 21 '18 at 7:15

@Jevgeni Geurtsen think the special relationship list AssignableFrom exist. Oh but it is most definitely there in the framework code, so the relation exists, but as long as it doesn't have a concrete implementation, its useless, as John Wu pointed out.
– Jevgeni Geurtsen
Nov 21 '18 at 8:16

add a comment |

If you have a type definition in mind and you want to do a type-compatibility check, just use <object> (or a suitable type if there is a type constraint) instead of <>:

var type1 = typeof(List<object>);

var type2 = typeof(IEnumerable<object>);



//returns true

type2.IsAssignableFrom(type1);

edited Nov 20 '18 at 11:46

vaxquis

7,51453757

answered Nov 20 '18 at 8:42

John Wu

29.4k42652

Thanks for your answer. It is a brief check method. I can understand type2.IsAssignableFrom(type1) behavior. But do you mean two open generic type are independent or no relationship between them before they become closed type or bounded?
– YonF
Nov 21 '18 at 5:32

Can you be specific about what you mean when you say "related?" Of course they have certain relationships, and not others.
– John Wu
Nov 21 '18 at 6:38

A relationship such as inheritance, implimentation but not inheritance and implimentation. But I am not sure whether any kind relatioinship exist between open generic type themselves. @Jevgeni Geurtsen think the special relationship list AssignableFrom exist.
– YonF
Nov 21 '18 at 7:07

Believe it or not, generic type definitions do not have any implementation. The implementation is compiled only when you specify the type parameter, and only for that type (although implementations for reference types can be re-used because all pointers are the same size, but that is just an implementation detail). So you see to ask whether List<> implements IEnumerable<> doesn't make much sense because it has no implementation to begin with.
– John Wu
Nov 21 '18 at 7:15

@Jevgeni Geurtsen think the special relationship list AssignableFrom exist. Oh but it is most definitely there in the framework code, so the relation exists, but as long as it doesn't have a concrete implementation, its useless, as John Wu pointed out.
– Jevgeni Geurtsen
Nov 21 '18 at 8:16

add a comment |

If you have a type definition in mind and you want to do a type-compatibility check, just use <object> (or a suitable type if there is a type constraint) instead of <>:

var type1 = typeof(List<object>);

var type2 = typeof(IEnumerable<object>);



//returns true

type2.IsAssignableFrom(type1);

edited Nov 20 '18 at 11:46

vaxquis

7,51453757

answered Nov 20 '18 at 8:42

John Wu

29.4k42652

If you have a type definition in mind and you want to do a type-compatibility check, just use <object> (or a suitable type if there is a type constraint) instead of <>:

var type1 = typeof(List<object>);

var type2 = typeof(IEnumerable<object>);



//returns true

type2.IsAssignableFrom(type1);

edited Nov 20 '18 at 11:46

vaxquis

7,51453757

answered Nov 20 '18 at 8:42

John Wu

29.4k42652

edited Nov 20 '18 at 11:46

vaxquis

7,51453757

edited Nov 20 '18 at 11:46

vaxquis

7,51453757

edited Nov 20 '18 at 11:46

vaxquis

7,51453757

answered Nov 20 '18 at 8:42

John Wu

29.4k42652

answered Nov 20 '18 at 8:42

John Wu

29.4k42652

answered Nov 20 '18 at 8:42

John Wu

29.4k42652

Thanks for your answer. It is a brief check method. I can understand type2.IsAssignableFrom(type1) behavior. But do you mean two open generic type are independent or no relationship between them before they become closed type or bounded?
– YonF
Nov 21 '18 at 5:32

Can you be specific about what you mean when you say "related?" Of course they have certain relationships, and not others.
– John Wu
Nov 21 '18 at 6:38

A relationship such as inheritance, implimentation but not inheritance and implimentation. But I am not sure whether any kind relatioinship exist between open generic type themselves. @Jevgeni Geurtsen think the special relationship list AssignableFrom exist.
– YonF
Nov 21 '18 at 7:07

Believe it or not, generic type definitions do not have any implementation. The implementation is compiled only when you specify the type parameter, and only for that type (although implementations for reference types can be re-used because all pointers are the same size, but that is just an implementation detail). So you see to ask whether List<> implements IEnumerable<> doesn't make much sense because it has no implementation to begin with.
– John Wu
Nov 21 '18 at 7:15

@Jevgeni Geurtsen think the special relationship list AssignableFrom exist. Oh but it is most definitely there in the framework code, so the relation exists, but as long as it doesn't have a concrete implementation, its useless, as John Wu pointed out.
– Jevgeni Geurtsen
Nov 21 '18 at 8:16

add a comment |

Thanks for your answer. It is a brief check method. I can understand type2.IsAssignableFrom(type1) behavior. But do you mean two open generic type are independent or no relationship between them before they become closed type or bounded?
– YonF
Nov 21 '18 at 5:32

Can you be specific about what you mean when you say "related?" Of course they have certain relationships, and not others.
– John Wu
Nov 21 '18 at 6:38

A relationship such as inheritance, implimentation but not inheritance and implimentation. But I am not sure whether any kind relatioinship exist between open generic type themselves. @Jevgeni Geurtsen think the special relationship list AssignableFrom exist.
– YonF
Nov 21 '18 at 7:07

Believe it or not, generic type definitions do not have any implementation. The implementation is compiled only when you specify the type parameter, and only for that type (although implementations for reference types can be re-used because all pointers are the same size, but that is just an implementation detail). So you see to ask whether List<> implements IEnumerable<> doesn't make much sense because it has no implementation to begin with.
– John Wu
Nov 21 '18 at 7:15

@Jevgeni Geurtsen think the special relationship list AssignableFrom exist. Oh but it is most definitely there in the framework code, so the relation exists, but as long as it doesn't have a concrete implementation, its useless, as John Wu pointed out.
– Jevgeni Geurtsen
Nov 21 '18 at 8:16

Thanks for your answer. It is a brief check method. I can understand type2.IsAssignableFrom(type1) behavior. But do you mean two open generic type are independent or no relationship between them before they become closed type or bounded?
– YonF
Nov 21 '18 at 5:32

Can you be specific about what you mean when you say "related?" Of course they have certain relationships, and not others.
– John Wu
Nov 21 '18 at 6:38

A relationship such as inheritance, implimentation but not inheritance and implimentation. But I am not sure whether any kind relatioinship exist between open generic type themselves. @Jevgeni Geurtsen think the special relationship list AssignableFrom exist.
– YonF
Nov 21 '18 at 7:07

Believe it or not, generic type definitions do not have any implementation. The implementation is compiled only when you specify the type parameter, and only for that type (although implementations for reference types can be re-used because all pointers are the same size, but that is just an implementation detail). So you see to ask whether List<> implements IEnumerable<> doesn't make much sense because it has no implementation to begin with.
– John Wu
Nov 21 '18 at 7:15

@Jevgeni Geurtsen think the special relationship list AssignableFrom exist. Oh but it is most definitely there in the framework code, so the relation exists, but as long as it doesn't have a concrete implementation, its useless, as John Wu pointed out.
– Jevgeni Geurtsen
Nov 21 '18 at 8:16

add a comment |

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