Method constraint must be super of class generic
How can I do the following?
I have a method which I need to say its generic type must be the super of the classes generic type.
Here is an example:
class Logger : ILogger, IStartable {}
///use
new FluentBuilder<Logger>().As<ILogger>().As<IStartable>();
This shows my intent, however does not work (as it is not syntactically incorrect):
public class FluentBuilder<TService> where TService : class
{
public FluentBuilder<TService> As<TContract>() where TService : TContract
{
return this;
}
}
c# .net generics
asked Nov 18 '18 at 11:52
dbonesdbones
3,02222644
add a comment |
How can I do the following?
I have a method which I need to say its generic type must be the super of the classes generic type.
Here is an example:
class Logger : ILogger, IStartable {}
///use
new FluentBuilder<Logger>().As<ILogger>().As<IStartable>();
This shows my intent, however does not work (as it is not syntactically incorrect):
public class FluentBuilder<TService> where TService : class
{
public FluentBuilder<TService> As<TContract>() where TService : TContract
{
return this;
}
}
c# .net generics
asked Nov 18 '18 at 11:52
dbonesdbones
3,02222644
Still it's unclear to me; why would you needAs<Ilogger>andAs<IStartable>(at the same time)?
– Stefan
Nov 18 '18 at 12:02
@Stefan updated my example, re the 2 As statements, it is an example consider Logger implements 2 interfaces (that we are interested in) and we want to describe that with the As<TContract>() method
– dbones
Nov 18 '18 at 12:56
add a comment |
How can I do the following?
I have a method which I need to say its generic type must be the super of the classes generic type.
Here is an example:
class Logger : ILogger, IStartable {}
///use
new FluentBuilder<Logger>().As<ILogger>().As<IStartable>();
This shows my intent, however does not work (as it is not syntactically incorrect):
public class FluentBuilder<TService> where TService : class
{
public FluentBuilder<TService> As<TContract>() where TService : TContract
{
return this;
}
}
c# .net generics
asked Nov 18 '18 at 11:52
dbonesdbones
3,02222644
How can I do the following?
I have a method which I need to say its generic type must be the super of the classes generic type.
Here is an example:
class Logger : ILogger, IStartable {}
///use
new FluentBuilder<Logger>().As<ILogger>().As<IStartable>();
This shows my intent, however does not work (as it is not syntactically incorrect):
public class FluentBuilder<TService> where TService : class
{
public FluentBuilder<TService> As<TContract>() where TService : TContract
{
return this;
}
}
c# .net generics
c# .net generics
asked Nov 18 '18 at 11:52
dbonesdbones
3,02222644
asked Nov 18 '18 at 11:52
dbonesdbones
3,02222644
edited Nov 18 '18 at 12:53
dbones
asked Nov 18 '18 at 11:52
dbonesdbones
3,02222644
asked Nov 18 '18 at 11:52
dbonesdbones
3,02222644
asked Nov 18 '18 at 11:52
dbonesdbones
3,02222644
3,02222644
Still it's unclear to me; why would you needAs<Ilogger>andAs<IStartable>(at the same time)?
– Stefan
Nov 18 '18 at 12:02
@Stefan updated my example, re the 2 As statements, it is an example consider Logger implements 2 interfaces (that we are interested in) and we want to describe that with the As<TContract>() method
– dbones
Nov 18 '18 at 12:56
add a comment |
Still it's unclear to me; why would you needAs<Ilogger>andAs<IStartable>(at the same time)?
– Stefan
Nov 18 '18 at 12:02
@Stefan updated my example, re the 2 As statements, it is an example consider Logger implements 2 interfaces (that we are interested in) and we want to describe that with the As<TContract>() method
– dbones
Nov 18 '18 at 12:56
Still it's unclear to me; why would you need
As<Ilogger> and As<IStartable> (at the same time)?– Stefan
Nov 18 '18 at 12:02
Still it's unclear to me; why would you need
As<Ilogger> and As<IStartable> (at the same time)?– Stefan
Nov 18 '18 at 12:02
@Stefan updated my example, re the 2 As statements, it is an example consider Logger implements 2 interfaces (that we are interested in) and we want to describe that with the As<TContract>() method
– dbones
Nov 18 '18 at 12:56
@Stefan updated my example, re the 2 As statements, it is an example consider Logger implements 2 interfaces (that we are interested in) and we want to describe that with the As<TContract>() method
– dbones
Nov 18 '18 at 12:56
add a comment |
1 Answer
1
active
oldest
votes
This is not exactlywhat you want, but maybe a step in the correct direction. You can implement your As method as extension method:
public static class Ex
{
public static FluentBuilder<TService> As<TService, TContract>(this FluentBuilder<TService> that)
where TContract : class
where TService : class, TContract
{
return that;
}
}
The usage syntax is then:
new FluentBuilder<Logger>().As<Logger, ILogger>().As<Logger, IStartable>();
answered Nov 18 '18 at 12:10
Klaus GütterKlaus Gütter
2,3781321
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is not exactlywhat you want, but maybe a step in the correct direction. You can implement your As method as extension method:
public static class Ex
{
public static FluentBuilder<TService> As<TService, TContract>(this FluentBuilder<TService> that)
where TContract : class
where TService : class, TContract
{
return that;
}
}
The usage syntax is then:
new FluentBuilder<Logger>().As<Logger, ILogger>().As<Logger, IStartable>();
answered Nov 18 '18 at 12:10
Klaus GütterKlaus Gütter
2,3781321
add a comment |
This is not exactlywhat you want, but maybe a step in the correct direction. You can implement your As method as extension method:
public static class Ex
{
public static FluentBuilder<TService> As<TService, TContract>(this FluentBuilder<TService> that)
where TContract : class
where TService : class, TContract
{
return that;
}
}
The usage syntax is then:
new FluentBuilder<Logger>().As<Logger, ILogger>().As<Logger, IStartable>();
answered Nov 18 '18 at 12:10
Klaus GütterKlaus Gütter
2,3781321
add a comment |
This is not exactlywhat you want, but maybe a step in the correct direction. You can implement your As method as extension method:
public static class Ex
{
public static FluentBuilder<TService> As<TService, TContract>(this FluentBuilder<TService> that)
where TContract : class
where TService : class, TContract
{
return that;
}
}
The usage syntax is then:
new FluentBuilder<Logger>().As<Logger, ILogger>().As<Logger, IStartable>();
answered Nov 18 '18 at 12:10
Klaus GütterKlaus Gütter
2,3781321
This is not exactlywhat you want, but maybe a step in the correct direction. You can implement your As method as extension method:
public static class Ex
{
public static FluentBuilder<TService> As<TService, TContract>(this FluentBuilder<TService> that)
where TContract : class
where TService : class, TContract
{
return that;
}
}
The usage syntax is then:
new FluentBuilder<Logger>().As<Logger, ILogger>().As<Logger, IStartable>();
answered Nov 18 '18 at 12:10
Klaus GütterKlaus Gütter
2,3781321
answered Nov 18 '18 at 12:10
Klaus GütterKlaus Gütter
2,3781321
answered Nov 18 '18 at 12:10
Klaus GütterKlaus Gütter
2,3781321
answered Nov 18 '18 at 12:10
Klaus GütterKlaus Gütter
2,3781321
2,3781321
add a comment |
add a comment |
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Still it's unclear to me; why would you need
As<Ilogger>andAs<IStartable>(at the same time)?– Stefan
Nov 18 '18 at 12:02
@Stefan updated my example, re the 2 As statements, it is an example consider Logger implements 2 interfaces (that we are interested in) and we want to describe that with the As<TContract>() method
– dbones
Nov 18 '18 at 12:56