Oye pero en ninguna de las carpetas que creaste tienes que ejecutar el comando de la creacion del ambiente? Yo ahi me perdi...
@linuxking88453 ай бұрын
encryption
@johanortega54654 ай бұрын
funciona?
@Insta360Travel5 ай бұрын
Me gusto mucho el contenido
@for-devs-com5 ай бұрын
Agradezco tu comentario, es gratificante saber que te ha gustado, esperamos seguir creciendo en el canal y mejorar nuestro contenido.
@francoeliaslanuzacenteno5 ай бұрын
Great Mix Bro'!!! 🙌🏻🙏🏻🐾
@for-devs-com21 күн бұрын
thanks!
@AshokKumar-gn1jy11 ай бұрын
Error
@for-devs-com11 ай бұрын
Hello Dev! Great effort on the puzzle - these can be really tricky sometimes! The output will actually be "Java", but also an "error". It's a common misconception, but in Java, String objects are immutable, meaning once they're created, they can't be changed. When you use `concat()`, it doesn't alter the original string; it creates a new one. So, for "JavaRocks" to be printed, you'd need to reassign the result back to `s` like so: ```java´´´ String s = "Java"; s = s.concat("Rocks"); System.out.println(s);
@dzvsow2643 Жыл бұрын
I think it might be Valid code. since the object is sub_class object and reference type is super_class. Am I wrong?
@for-devs-com Жыл бұрын
Hello @dzvsow2643 Thanks for your comment. You're right! In Java, a superclass reference can indeed point to a subclass object. This is a key aspect of polymorphism in object-oriented programming. Just remember, the superclass reference will only access methods and fields defined in the superclass, unless they're overridden in the subclass. Great observation, keep exploring Java!