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Home/ Questions/Q 8050691
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T07:08:19+00:00 2026-06-05T07:08:19+00:00

It seems like the C# compiler infers types differently depending on how a method

  • 0

It seems like the C# compiler infers types differently depending on how a method is called:

void Foo<T>() where T : Bar
{
   var instance = new T()
   {
      ID = 1
   }.
   ExtensionMethod();
}

In this case the compiler seems to infer that the type of instance is Bar, because I have a class Bar where ExtensionMethod is declared.

void Foo<T>() where T : Bar
{
   var instance = new T()
   {
      ID = 1
   };
   instance.ExtensionMethod();
}

In this case the compiler infers that the type of instance is T, which is what I would expect it to do in the first case as well.
Why is there such a difference?

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  1. Editorial Team
    Editorial Team
    2026-06-05T07:08:20+00:00Added an answer on June 5, 2026 at 7:08 am

    In the first case, you assign the result of the method call to instance. In the second case you discard the call’s result. Instead, you assign the new T This is the only difference.

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