Immutable + goto?

[Max Samukha]

On Thu, 19 Mar 2009 16:58:27 +0100, Frits van Bommel
<fvbommel at REMwOVExCAPSs.nl> wrote:

>dsimcha wrote:
>> import std.stdio;
>> 
>> uint bar = 0;
>> 
>> void main() {
>>     start:
>>     immutable uint foo = bar;
>>     bar++;
>>     writeln(foo);
>>     goto start;
>> }
>> 
>> foo changes in this case.  Is this a real bug, or is it considered undefined
>> behavior to use goto in this way?
>
>I disagree: foo doesn't change there :?.
>
>Declaring a local variable like that actually creates a new scope behind 
>the scenes, containing everything up to '}' (taking nesting into 
>account, of course). This means the label is outside the scope of foo, 
>and foo gets "destroyed" when the goto jumps out of it. A "new" foo is 
>created (with a different value) when the program enters its scope.
>
>This is basically the same as:
>-----
>while (true) {
>     immutable uint foo = bar;
>     bar++;
>     writefln(foo);
>}
>-----
>or even:
>-----
>while (true) {
>     void nested_fn(uint foo) {
>         bar++;
>         writefln(foo);
>     }
>     nested_fn(bar);
>}
>-----
>which show the behavior more clearly.

Then this is a really weird behavior:

void main()
{
start:
    //int j = bar;
    scope(exit)
        writefln("exit");
    writefln(bar);
    bar++;
    if (bar == 10)
        return;
    goto start;
}

scope(exit) is not called at all when the local is commented out and
called 10 times, otherwise. This implicit scope looks more like a bug
or misfeature to me. Not calling scope(exit) if there is no variable
declared is definitely a bug.