[HD-OCP17/21-Fundamentals Pg 242, Sec. 10.2.1 - using-instanceof-as-a-pattern-matching-operator]

[raphaelzintec]

void main(String[] args){
Object obj = getObject(args.length);
if(! (obj instanceof String str) ){
//will not compile if return statement is commented out
return; //or a throw an exception!
}
System.out.println("str = "+str);
}

this out of scope has no real logic, it works only with ! operator and empty return.
Why i mean by that is that compiler doesnt care if its true or false, it will be correct

[pheroidsmok]

void main(String[] args){
Object obj = getObject(args.length)geometry dash;
if(! (obj instanceof String str) ){
//will not compile if return statement is commented out
return; //or a throw an exception!
}
System.out.println("str = "+str);
}

this out of scope has no real logic, it works only with ! operator and empty return.
Why i mean by that is that compiler doesnt care if its true or false, it will be correct

Sorry, but I still don't really understand this code. How does the return statement work?

[admin]

The return statement ensures that there is no path left in the code that the control can take when obj is not an instance of String. Therefore, the compiler knows for sure that str will be in scope everywhere else. This is why str is available outside the if block.

Without the return statement, the print statement will be executed even when !(obj instanceof String str) is false. But in that case str will not be in flow scope and so the compiler refuses to accept this code.