Answer 1

Koloroj, you changed the question. Before you were saying that the while loop is repeating forever when you change `int namE` to `char namE`. That's why I explained in terms of char and int.

Anyways, the problem is, as Martin Taylor said, that if you enter an invalid input (example: "Bananas" where integer is needed), the input operation would simply fail, but the program would continue without any errors. That is why you need to check after each input if the input operation failed, and if it did, perform the necessary actions. This has been demonstrated by Martin Taylor in his answer.

Answer 2

Firstly, please save the code in a C++ code in the code playground and then link it here. It makes it easier to reference line numbers and errors/warnings.

So you changed the type of `namE` to char, and then gave the second input (input into `namE`) as "1" (or "2" withtout the quotes), right?

The first thing for you to know here is that char is actually an integral type, that is
char character = 'A';
here, `character` actually holds 65, which is the ordinal value of the character 'A'.

So when you give the input '1' (or '2') for variable `namE`, the value in `namE` will actually be 49 or 50.

But as you can see on line 16, 21 and 24, you are comparing `nameE` with values like 1 and 2, which won't give error as `namE` is also actually an integer, but will always be false.

So the simplest fix would be to have `namE` of type int, but if you want it to be char, you need to change the 1s ans 2s on line 16, 21 and 24 to '1's and '2's (the single quotes will make it a char which can then be compared to `namE`)