How can I prevent DRY for this method?

Question

Having each scope almost the same like below doesn't feel right at all. I'd rather have one scope where name, age i.e. are set than having 3 for each object of cat, dog and puppy. How can this code below be improved? private void addAnimal_Click( object sender, EventArgs e) { var name = inputName.Text; int age = int .Parse(inputAge.Text); var favfood = inputFavfood.Text; var breed = inputBreed.Text; var typeOfAnimal = dropDownTypeOfAnimal.Text; if (typeOfAnimal == "Cat" ) { Animal cat = new Cat(); cat.Name = name; cat.Age = age; cat.FavFood = favfood; cat.Breed = breed; cat.IsHungry = utility.ShuffleProbability(); Pets.Add(cat); } if (typeOfAnimal == "Dog" ) { Animal dog = new Dog(); dog.Name = name; dog.Age = age; dog.FavFood = favfood; dog.Breed = breed; dog.IsHungry = utility.ShuffleProbability(); Pets.Add(dog); } if (typeOfAnimal == "Puppy" ) { Animal puppy = new Puppy(); puppy.Name = name; puppy.Age = age; puppy.FavFood = favfood; puppy.Breed = breed; puppy.IsHungry = utility.ShuffleProbability(); Pets.Add(puppy); } }

Sachin Singh · Answer

you can directly create the object based on string and in this way you will be free from if else. private void addAnimal_Click( object sender, EventArgs e) { var name = inputName.Text; int age = int .Parse(inputAge.Text); var favfood = inputFavfood.Text; var breed = inputBreed.Text; var typeOfAnimal = dropDownTypeOfAnimal.Text; var assemblyName = "ConsoleTest" ; // your namespace name Type t = Type.GetType(assemblyName+ "." +typeOfAnimal ); var obj= Activator.CreateInstance(t) as typeOfAnimal ; obj.Name = name; obj.Age = age; obj.FavFood = favfood; obj.Breed = breed; obj.IsHungry = utility.ShuffleProbability(); Pets.Add(cat); } } Note :- I have used Reflection.

Rijwan Ansari · Answer

In your case, I don't see any different properties based on if condition except object itself. Simply, private void addAnimal_Click( object sender, EventArgs e) { var name = inputName.Text; int age = int .Parse(inputAge.Text); var favfood = inputFavfood.Text; var breed = inputBreed.Text; var typeOfAnimal = dropDownTypeOfAnimal.Text; Animal animal = null ; if (typeOfAnimal == "Cat" ) { animal = new Cat(); } else if (typeOfAnimal == "Dog" ) { animal = new Dog(); } else if (typeOfAnimal == "Puppy" ) { animal = new Puppy(); } animal.Name = name; animal.Age = age; animal.FavFood = favfood; animal.Breed = breed; animal.IsHungry = utility.ShuffleProbability(); Pets.Add(animal); } Addtionally, use if else or switch for better, as it will skip other conditions if one case is matched. Regards

Kiran Mohanty · Answer

Animal animal= null ; if (typeOfAnimal == "cat" ){ animal = new Cat(); } elseif(typeOfAnimal == "dog" ){ animal = new Dog(); } else if (typeOfAnimal == "puppy" ){ animal = new Puppy(); } else { animal = new Animal(); } animal.Name = name; animal.Age = age; // initialize all properties

Laxmidhar Sahoo · Answer

hi No need to take three classes just use one common calss named like petanimal and a distingushed filed like pet type and add with the list