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Local Meaning of Union Types (safe assignments)  #17188

Description

@olegdunkan

TypeScript Version: Playground

Code

class A {
    x: number;
}

class B extends A {
    y: number;
}

class C extends A {
    z: number;
}

declare function process(c:C);
function f(x:  B | C) {
    
    if (x instanceof C) {
        //we memorize x in c
        let c = x; //here x of type C        
      
        //restore x
        x = c;  //here x loses the narrowed type and x is now B | C again and is's by design    
        x.z = 1; //no error, what? 
        process(x);  //it have to be an error, what?
    }
}

Preamble:
A type guard for a variable x has no effect if the statements or expressions it guards contain assignments to x.

If compiler knows that the type of right hand expression is the same as narrowed type of parameter (in my case) why does compiler expand type of x to the initial type? Maybe it is possible to check for this type of assignments?

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