Why destructor is called when going backward via goto [duplicate]

Question

Possible Duplicate:
Will using goto leak variables?

In the following example, when goto is called to go "backwards", the destructor of A is called. Why is it like that? Object a is not leaving its scope, is it? Does the standard say anything about this behavior regarding goto?

void f() { start:     A a;     goto start; } 

Answer 1

6.6 Jump statements [stmt.jump]

Paragraph 2:

On exit from a scope (however accomplished), objects with automatic storage duration (3.7.3) that have been constructed in that scope are destroyed in the reverse order of their construction. [ Note: For temporaries, see 12.2. —end note] Transfer out of a loop, out of a block, or back past an initialized variable with automatic storage duration involves the destruction of objects with automatic storage duration that are in scope at the point transferred from but not at the point transferred to. (See 6.7 for transfers into blocks). [Note: However, the program can be terminated (by calling std::exit() or std::abort() (18.5), for example) without destroying class objects with automatic storage duration. — end note ]

I think the important part is:

or back past an initialized variable with automatic storage duration involves the destruction

Answer 2

The lifetime of object a starts from its declaration, and extends to the end of the block containing it.

This means that, in jumping back to before the declaration, you jump to a stack frame situations where the local didn't exist, so it must be destructed

1. The point of declaration for a name is immediately after its complete declarator (Clause 8) and before its initializer (if any) , [...] (§ 3.3.2)

2. A name declared in a block (6.3) is local to that block; it has block scope. Its potential scope begins at its point of declaration (3.3.2) and ends at the end of its block. A variable declared at block scope is a local variable. (§ 3.3.3)