c# - How to return an implementation of an interface with interface as return type? -

i have interface:isearch<t> , have class implements interface: filesearch : isearch<filesearchresult>

i have class filesearchargs : searchargs has method return search object:

public override isearch<searchresult> getsearchobject ()  {     return ((isearch<searchresult>)new filesearch());  } 

this overridden from:

public virtual isearch<searchresult> getsearchobject () { return null; } 

the code build if provided cast (isearch), throws exception @ runtime unable cast error. additionally, previous iteration did not apply generics interface, , method signature of getsearchobject() follows:

public override isearch getsearchobject() { return new filesearch();} 

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i know 1 solution return base class "search" instead of implementation of interface, i'd prefer not this, , understand why cannot follow pattern had.

any help, appreciated. i'm trying simplify going on, if clarification needed, please let me know!

thanks in advance.

try declare interface this:

interface isearch<out t> {    // ... } 

(assuming filesearchresult inherits searchresult, , type parameter occurs in covariant positions in interface)

or if you'll use searchresults' children:

interface isearch<out t> t : searchresult {    // ... } 

update:

now know use type parameter in input position, use base non-generic interface:

interface isearch { } interface isearch<t> : isearch t : searchresult { }  // ...  public isearch getsearchobject() {    return new filesearch();  }  

or segregate interfaces (pdf) (if makes sense you):

interface isearchco<out t> t : searchresult {   t result { get; } } interface isearchcontra<in t> t : searchresult {   t result { set; } }  // ...  public isearchco<searchresult> getsearchobject() {    return (isearchco<searchresult>)new filesearch();  }