C# switch variable initialization: Why does this code NOT cause a compiler error or a runtime error?

Question

...
case 1:
   string x = \"SomeString\";
   ...
   break;
case 2:
   x = \"SomeOtherString\";
   ...
   break;
...

Is there something that

Answer 1

There is no compiler error because the switch statement does not create a new scope for variables.

If you declare a variable inside of a switch, the variable is in the same scope as the code block surrounding the switch. To change this behavior, you would need to add {}:

...
case 1:
    // Start a new variable scope 
    {
        string x = "SomeString";
        ...
    }
    break;
case 2:
    {
        x = "SomeOtherString";
        ...
    }
    break;
...

This will cause the compiler to complain. However, switch, on it's own, doesn't internally do this, so there is no error in your code.