The version with int r = x + y;
won’t compile either.
The problem is that it is possible for r
to come to scope without its initializer being executed. The code would compile fine if you removed the initializer completely (i.e. the line would read int r;
).
The best thing you can do is to limit the scope of the variable. That way you’ll satisfy both the compiler and the reader.
switch(i)
{
case 1:
{
int r = 1;
cout << r;
}
break;
case 2:
{
int r = x - y;
cout << r;
}
break;
};
The Standard says (6.7/3):
It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps from a point where a local variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has POD type (3.9) and is declared without an initializer (8.5).