Question

Why does this code fragment run fine

void foo(int i)
{
    switch(i) {
    case 1:
    {
        X x1;
        break;
    }
    case 2:
        X x2;
        break;
    }
}

whereas the following gives compilation error (initialization of ‘x1’ is skipped by ‘case’ label)?

void foo(int i)
{
    switch(i) {
    case 1:
        X x1;
        break;
    case 2:
        X x2;
        break;
    }
}

I understand that using braces introduces a new scope, hence storage will not be allocated for x1 till we hit its opening brace. But x2 is still initialized inside a case label without enclosing braces. Should this not be an error too?

I think initialization of x2 can be conditionally skipped in both the code fragments

Answer 1

1: Valid

 case 1:
        {
        X x1;
        break;
        }

If it doesn’t hit the condition, x1 can’t be used by any further statements, so there can’t be a runtime error with this. x1 doesn’t try to exist outside braces.

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2: Invalid

 switch(i) {
    case 1:
        X x1; //don't break
        i = 2;
        ...
        ...
        ...
     case 2:
        x1.someOperation()

 }

In the above, if i was 2 initially, you’d hit x1.someOperation() before X x1 which would construct the object.

If it was allowed to compile, it would throw a runtime error or not, depending upon whether the case:1 was executed before 2, (and the object was constructed). Hence, it is disallowed by the compiler.

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The same is allowed with Plain Old Data types which cannot have a user-defined constructor.