Question

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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