Cannot implement generic type guarding function

TypeScript Version: 2.5.2

Code

interface ITypeChecker
{
    isOfType<T extends TBase, TBase>(obj: TBase): obj is T; 
}


class Checker implements ITypeChecker
{
    isOfType<T extends TBase, TBase>(obj: TBase): obj is T
    {
        return true;
    }

}

Expected behavior:
Class Checker should be treated as a valid implementation (used to be in Ts-2.3).
Actual behavior:
Compiler complains:

interface-not-properly-implemented
(...)
- Type predicate 'value is TBase' is not assignable to 'value is T'
  - Type 'TBase' is not assignable to type 'T'

When comparing two generic signatures, the source is instantiated in the context of the target then both are compared (see #16368 for more details). since the generic type parameter T is never used in your signature in an input position, so it will not be instantiated, and as a result when comparing the two signatures, T from the first is not the same T from the second, and the comparison will fail, since the compiler can not make any assumptions about the relationship of the two Ts.

The core of the issue here is that is equivalent to casting. i would argue, you are better off casting and avoiding the runtime cost of a function that can not verify the type anyways.

Alternatively, just use --noStrictGenericChecks to disable checking the generic signatures.

The other reason this does not work is that type guards are not an inference position. now that we are inferring from return type position, seems like we should do this too.

No longer errors

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Cannot implement generic type guarding function #18202

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Cannot implement generic type guarding function #18202

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mhegazy commented on Sep 5, 2017

mhegazy commented on Sep 11, 2017

RyanCavanaugh commented on Jul 23, 2025

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