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118 i found the example above confusing Because under the hood, the compiler will go away and create a new type (sometimes called a closed generic type) for each different usage of the open generic type I am using react and jsx so i think it complicated the scenario
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I got clarification from typescript deep dive, which states for arrow generics I am not sure if it is possible for primitive types and how if so. Use extends on the generic parameter to hint the compiler that it's a generic, this came from a simpler example that helped me.
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You can certainly define generic delegates, after all, that's exactly what func and action are
They are treated as generic definitions, just like generic interfaces and classes are However, you cannot use generic definitions in method signatures, only parameterized generic types Quite simply you cannot do what you are trying to achieve with a delegate alone. What's the best way to call a generic method when the type parameter isn't known at compile time, but instead is obtained dynamically at runtime
Why do we observe this weird behaviour What keeps us from comparing the values of generic types which are known to be icomparable Doesn't it somehow defeat the entire purpose of generic constraints How do i resolve this, or at least work around it?
The generic parameter type will be the same for all methods, so i would like it at the class level
I know i could make a generic version and then inherit from it for the int version, but i was just hoping to get it all in one.but i didn't know of any way to do that. I have a generics class, foo<t> In a method of foo, i want to get the class instance of type t, but i just can't call t.class What is the preferred way to get around it using t.class?
I have the following method with generic type I would like to limit t to primitive types such as int, string, float but not class type I know i can define generic for class type like this
